Class eight math solution

বীজগণিতীয় ভগ্নাংশ

অনুশীলনী ৫.১ এর সমাধান

১। লঘিষ্ঠ আকারে প্রকাশ কর:
(ক) \frac{{4{x^2}{y^3}{z^5}}}{{9{x^5}{y^2}{z^3}}}            (খ) \frac{{16{{(2x)}^4}{{(3y)}^5}}}{{{{(3x)}^3}.{{(2y)}^6}}}

(গ) \frac{{{x^3}y + x{y^3}}}{{{x^2}{y^3} + {x^3}{y^2}}}         (ঘ)  \frac{{(a - b)(a + b)}}{{{a^3} - {b^3}}}

(ঙ) \frac{{{x^2} - 6x + 5}}{{{x^2} - 25}}          (চ)\frac{{{x^2} - 7x + 12}}{{{x^2} - 9x + 20}}

(ছ)  \frac{{({x^3} - {y^3})({x^2} - xy + {y^2})}}{{({x^2} - {y^2})({x^3} + {y^3})}}         (জ)\frac{{{a^2} - {b^2} - 2bc - {c^2}}}{{{a^2} + 2ab + {b^2} - {c^2}}}

সমাধান ১(ক):
প্রদত্ত ভগ্নাংশ = \frac{{4{x^2}{y^3}{z^5}}}{{9{x^5}{y^2}{z^3}}}

= \frac{{4.{x^2}.y.{y^2}.{z^3}.{z^2}}}{{9.{x^2}.{x^3}.{y^2}.{z^3}}}

= \frac{{4y{z^2}}}{{9{x^3}}}    (Ans.)

সমাধান ১(খ):
প্রদত্ত ভগ্নাংশ =\frac{{16{{(2x)}^4}{{(3y)}^5}}}{{{{(3x)}^3}.{{(2y)}^6}}}

= \frac{{{2^4}{{.2}^4}.{x^4}{{.3}^5}.{y^5}}}{{{3^3}.{x^3}{{.2}^6}.{y^6}}}

= \frac{{{2^8}{{.3}^5}{x^4}.{y^5}}}{{{2^6}{{.3}^3}.{x^3}.{y^6}}}

= \frac{{{2^2}{{.3}^2}.x}}{y}

= \frac{{36x}}{y}    (Ans.)          

সমাধান ১(গ):
প্রদত্ত ভগ্নাংশ =\frac{{{x^3}y + x{y^3}}}{{{x^2}{y^3} + {x^3}{y^2}}}

= \frac{{xy({x^2} + {y^2})}}{{{x^2}{y^2}(y + x)}}

= \frac{{{x^2} + {y^2}}}{{xy(x + y)}}   (Ans.)

সমাধান ১(ঘ):
প্রদত্ত ভগ্নাংশ= \frac{{(a - b)(a + b)}}{{{a^3} - {b^3}}}

= \frac{{(a - b)(a + b)}}{{(a - b)({a^2} + ab + {b^2})}}

= \frac{{a + b}}{{{a^2} + ab + {b^2}}}     (Ans.)

সমাধান ১(ঙ):
প্রদত্ত ভগ্নাংশ= \frac{{{x^2} - 6x + 5}}{{{x^2} - 25}}

= \frac{{{x^2} - 5x - x + 5}}{{{x^2} - {5^2}}}

= \frac{{x(x - 5) - 1(x - 5)}}{{(x + 5)(x - 5)}}

= \frac{{(x - 5)(x - 1)}}{{(x + 5)(x - 5)}}

= \frac{{x - 1}}{{x + 5}}    (Ans.)

সমাধান ১(চ):
প্রদত্ত ভগ্নাংশ = \frac{{{x^2} - 7x + 12}}{{{x^2} - 9x + 20}}

= \frac{{{x^2} - 4x - 3x + 12}}{{{x^2} - 4x - 5x + 20}}

= \frac{{x(x - 4) - 3(x - 4)}}{{x(x - 4) - 5(x - 4)}}

= \frac{{(x - 4)(x - 3)}}{{(x - 4)(x - 5)}}

= \frac{{x - 3}}{{x - 5}}     (Ans.)

সমাধান ১(ছ):
প্রদত্ত ভগ্নাংশ= \frac{{({x^3} - {y^3})({x^2} - xy + {y^2})}}{{({x^2} - {y^2})({x^3} + {y^3})}}

= \frac{{(x - y)({x^2} + xy + {y^2})({x^2} - xy + {y^2})}}{{(x + y)(x - y)(x + y)({x^2} - xy + {y^2})}}

= \frac{{{x^2} + xy + {y^2}}}{{{{(x + y)}^2}}}     (Ans.)

সমাধান ১(জ):
প্রদত্ত ভগ্নাংশ= \frac{{{a^2} - {b^2} - 2bc - {c^2}}}{{{a^2} + 2ab + {b^2} - {c^2}}}

= \frac{{{a^2} - ({b^2} + 2bc + {c^2})}}{{({a^2} + 2ab + {b^2}) - {c^2}}}

= \frac{{{a^2} - {{(b + c)}^2}}}{{{{(a + b)}^2} - {c^2}}}

= \frac{{(a + b + c)(a - b - c)}}{{(a + b + c)(a + b - c)}}

= \frac{{a - b - c}}{{a + b - c}}     (Ans.)

২। সাধারণ হরবিশিষ্ট ভগ্নাংশে প্রকাশ কর:
(ক) \frac{{{x^2}}}{{xy}},\frac{{{y^2}}}{{yz}},\frac{{{z^2}}}{{zx}}     (খ) \frac{{x - y}}{{xy}},\frac{{y - z}}{{yz}},\frac{{z - x}}{{zx}}

(গ) \frac{x}{{x - y}},\frac{y}{{x + y}},\frac{z}{{x(x + y)}}       (ঘ) \frac{{x + y}}{{{{(x - y)}^2}}},\frac{{x - y}}{{{x^3} + {y^3}}},\frac{{y - z}}{{{x^2} - {y^2}}}

(ঙ) \frac{a}{{{a^3} + {b^3}}},\frac{b}{{{a^2} + ab + {b^2}}},\frac{c}{{{a^3} - {b^3}}}
(চ) \frac{1}{{{x^2} - 5x + 6}},\frac{1}{{{x^2} - 7x + 12}},\frac{1}{{{x^2} - 9x + 20}}

(ছ) \frac{{a - b}}{{{a^2}{b^2}}},\frac{{b - c}}{{{b^2}{c^2}}},\frac{{c - a}}{{{c^2}{a^2}}}     (জ)  \frac{{x - y}}{{x + y}},\frac{{y - z}}{{y + z}},\frac{{z - x}}{{z + x}}

সমাধান ২(ক):
\frac{{{x^2}}}{{xy}},\frac{{{y^2}}}{{yz}},\frac{{{z^2}}}{{zx}}
এখানে, ১ম ভগ্নাংশের হর = {xy}
২য় ভগ্নাংশের হর ={yz}
৩য় ভগ্নাংশের হর ={zx}
\therefore হরগুলোর ল.সা.গু = xyz
\therefore \frac{{{x^2}}}{{xy}}
= \frac{{{x^2}}}{{xy}} \times \frac{z}{z}    [\because xyz \div xy = z]
= \frac{{{x^2}z}}{{xyz}}

এখন, \frac{{{y^2}}}{{yz}}
= \frac{{{y^2}}}{{yz}} \times \frac{x}{x}     [\because xyz \div yz = x]
= \frac{{x{y^2}}}{{xyz}}

এবং \frac{{{z^2}}}{{zx}}
= \frac{{{z^2}}}{{zx}} \times \frac{y}{y}    [\because xyz \div zx = y]
= \frac{{y{z^2}}}{{xyz}}
\therefore সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
   \frac{{{x^2}z}}{{xyz}}, \frac{{{y^2}x}}{{xyz}}, \frac{{{z^2}.y}}{{xyz}}    (Ans.)
 

সমাধান ২(খ):
\frac{{x - y}}{{xy}},\frac{{y - z}}{{yz}},\frac{{z - x}}{{zx}}

এখানে, ১ম ভগ্নাংশের হর = {xy}
২য় ভগ্নাংশের হর ={yz}
৩য় ভগ্নাংশের হর ={zx}
\therefore হরগুলোর ল.সা.গু = xyz
\therefore \frac{{x - y}}{{xy}}
= \frac{{x - y}}{{xy}} \times \frac{z}{z}    [\because xyz \div xy = z]
= \frac{{\left( {x - y} \right)z}}{{xyz}}

এখন, \frac{{y - z}}{{yz}}
= \frac{{y - z}}{{yz}} \times \frac{x}{x}    [\because xyz \div yz = x]
= \frac{{x(y - z)}}{{xyz}}

এবং \frac{{z - x}}{{zx}}
= \frac{{z - x}}{{zx}} \times \frac{y}{y}     [\because xyz \div zx = y]
= \frac{{y\left( {z - x} \right)}}{{xyz}}

\therefore সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
\frac{{(x - y)z}}{{xyz}}, \frac{{(y - z)x}}{{xyz}}, \frac{{(z - x)y}}{{xyz}}    (Ans.)

সমাধান ২(গ):
\frac{x}{{x - y}},\frac{y}{{x + y}},\frac{z}{{x(x + y)}}      
১ম ভগ্নাংশের হর { = x - y}
২য় ভগ্নাংশের হর { = x + y}
৩য় ভগ্নাংশের হর { = x(x + y)}
\therefore হরগুলোর ল.সা.গু = x(x + y)(x - y)
\therefore \frac{x}{{x - y}}
= \frac{x}{{x - y}} \times \frac{{x(x + y)}}{{x(x + y)}}   [\because x(x + y)(x - y) \div (x - y) = x(x + y)]

= \frac{{{x^2}(x + y)}}{{x(x + y)(x - y)}}

= \frac{{{x^2}(x + y)}}{{x({x^2} - {y^2})}}

এখন, \frac{y}{{x + y}}

= \frac{y}{{x + y}} \times \frac{{x(x - y)}}{{x(x - y)}}    [\because x(x + y)(x - y) \div (x + y) = x(x - y)]

= \frac{{xy(x - y)}}{{x({x^2} - {y^2})}}

এবং \frac{z}{{x(x + y)}}

= \frac{z}{{x(x + y)}} \times \frac{{x - y}}{{x - y}} [\because x(x + y)(x - y) \div x(x + y) = x - y]

= \frac{{z(x - y)}}{{x({x^2} - {y^2})}}

সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
\frac{{{x^2}(x + y)}}{{x({x^2} - {y^2})}}, \frac{{xy(x - y)}}{{x({x^2} - {y^2})}}, \frac{{z(x - y)}}{{x({x^2} - {y^2})}}    (Ans.)

সমাধান ২(ঘ):
\frac{{x + y}}{{{{(x - y)}^2}}},\frac{{x - y}}{{{x^3} + {y^3}}},\frac{{y - z}}{{{x^2} - {y^2}}}
এখানে, ১ম ভগ্নাংশের হর { = {{(x - y)}^2}}
২য় ভগ্নাংশের হর = {x^3} + {y^3} = (x + y)({x^2} - xy + {y^2})
৩য় ভগ্নাংশের হর = {x^2} - {y^2} = (x + y)(x - y)
\therefore হরগুলোর ল.সা.গু = {(x - y)^2}(x + y)({x^2} - xy + {y^2})

\therefore \frac{{x + y}}{{{{(x - y)}^2}}}

= \frac{{x + y}}{{{{(x - y)}^2}}} \times \frac{{(x + y)({x^2} - xy + {y^2})}}{{(x + y)({x^2} - xy + {y^2})}}   [\because {(x - y)^2}(x + y)({x^2} - xy + {y^2}) \div {(x - y)^2} = (x + y)({x^2} - xy + {y^2})]

= \frac{{(x + y)({x^3} + {y^3})}}{{{{(x - y)}^2}({x^3} + {y^3})}}

এখন, \frac{{x - y}}{{{x^3} + {y^3}}}
= \frac{{x - y}}{{{x^3} + {y^3}}} \times \frac{{{{(x - y)}^2}}}{{{{(x - y)}^2}}}    [\because {(x - y)^2}(x + y)({x^2} - xy + {y^2}) \div ({x^3} + {y^3}) = {(x - y)^2}]

= \frac{{{{(x - y)}^3}}}{{{{(x - y)}^2}({x^3} + {y^3})}}

এবং \frac{{y - z}}{{{x^2} - {y^2}}}

= \frac{{y - z}}{{{x^2} - {y^2}}} \times \frac{{(x - y)({x^2} - xy + {y^2})}}{{(x - y)({x^2} - xy + {y^2})}}    [\because {(x - y)^2}(x + y)({x^2} - xy + {y^2}) \div {x^2} - {y^2} = (x - y)({x^2} - xy + {y^2})]

= \frac{{(y - z)(x - y)({x^2} - xy + {y^2})}}{{(x + y)(x - y)(x - y)({x^2} - xy + {y^2})}}
= \frac{{(y - z)(x - y)({x^2} - xy + {y^2})}}{{{{(x - y)}^2}({x^3} + {y^3})}}    [\because (a + b)({a^2} - ab + {b^2}) = {a^3} + {b^3}]

সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
\frac{{(x + y)({x^3} + {y^3})}}{{{{(x - y)}^2}({x^3} + {y^3})}}, = \frac{{{{(x - y)}^3}}}{{{{(x - y)}^2}({x^3} + {y^3})}}, = \frac{{(y - z)(x - y)({x^2} - xy + {y^2})}}{{{{(x - y)}^2}({x^3} + {y^3})}}   

সমাধান ২(ঙ):
\frac{a}{{{a^3} + {b^3}}},\frac{b}{{{a^2} + ab + {b^2}}},\frac{c}{{{a^3} - {b^3}}}
এখানে, ১ম ভগ্নাংশের হর = {a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})
২য় ভগ্নাংশের হর ={{a^2} + ab + {b^2}}
৩য় ভগ্নাংশের হর = {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})
\therefore হরগুলোর ল.সা.গু = (a + b)(a - b)({a^2} + ab + {b^2})({a^2} - ab + {b^2})

অতএব, \frac{a}{{{a^3} + {b^3}}}

= \frac{a}{{{a^3} + {b^3}}} \times \frac{{(a - b)({a^2} + ab + {b^2})}}{{(a - b)({a^2} + ab + {b^2})}}    [\because (a + b)(a - b)({a^2} + ab + {b^2})({a^2} - ab + {b^2}) \div {a^3} + {b^3} = (a - b)({a^2} + ab + {b^2})]

= \frac{{a(a - b)({a^2} + ab + {b^2})}}{{({a^3} + {b^3})(a - b)({a^2} + ab + {b^2})}}
= \frac{{a(a - b)({a^2} + ab + {b^2})}}{{({a^3} + {b^3})({a^3} - {b^3})}}
= \frac{{a({a^3} - {b^3})}}{{({a^3} + {b^3})({a^3} - {b^3})}}

এখন, \frac{b}{{{a^2} + ab + {b^2}}}

= \frac{b}{{{a^2} + ab + {b^2}}} \times \frac{{(a - b)(a + b)({a^2} - ab + {b^2})}}{{(a - b)(a + b)({a^2} - ab + {b^2})}}     [\because (a + b)(a - b)({a^2} + ab + {b^2})({a^2} - ab + {b^2}) \div {a^2} + ab + {b^2} = (a - b)(a + b)({a^2} - ab + {b^2})]

= \frac{{b(a - b)(a + b)({a^2} - ab + {b^2})}}{{({a^2} + ab + {b^2})(a - b)(a + b)({a^2} - ab + {b^2})}}

= \frac{{b(a - b)({a^3} + {b^3})}}{{({a^3} - {b^3})({a^3} + {b^3})}}

এবং \frac{c}{{{a^3} - {b^3}}}

= \frac{c}{{{a^3} - {b^3}}} \times \frac{{(a + b)({a^2} - ab + {b^2})}}{{(a + b)({a^2} - ab + {b^2})}}    [\because (a + b)(a - b)({a^2} + ab + {b^2})({a^2} - ab + {b^2}) \div (a - b)({a^2} + ab + {b^2}) = (a + b)({a^2} - ab + {b^2})]

= \frac{{c(a + b)({a^2} - ab + {b^2})}}{{({a^3} - {b^3})(a + b)({a^2} - ab + {b^2})}}

= \frac{{c({a^3} + {b^3})}}{{({a^3} - {b^3})({a^3} + {b^3})}}

সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
\frac{{a(a - b)({a^2} + ab + {b^2})}}{{({a^3} + {b^3})({a^3} - {b^3})}}, \frac{{b(a - b)({a^3} + {b^3})}}{{({a^3} - {b^3})({a^3} + {b^3})}}, \frac{{c({a^3} + {b^3})}}{{({a^3} - {b^3})({a^3} + {b^3})}}

সমাধান ২(চ):
\frac{1}{{{x^2} - 5x + 6}},\frac{1}{{{x^2} - 7x + 12}},\frac{1}{{{x^2} - 9x + 20}}
এখানে, ১ম ভগ্নাংশের হর = {x^2} - 5x + 6
= {x^2} - 3x - 2x + 6
= x(x - 3) - 2(x - 3)
= (x - 3)(x - 2)

২য় ভগ্নাংশের হর = {x^2} - 7x + 12
= {x^2} - 4x - 3x + 12
= x(x - 4) - 3(x - 4)
= (x - 4)(x - 3)

৩য় ভগ্নাংশের হর = {x^2} - 9x + 20
= {x^2} - 4x - 5x + 20
= x(x - 4) - 5(x - 4)
= (x - 4)(x - 5)

\therefore হরগুলোর ল.সা.গু = (x - 2)(x - 3)(x - 4)(x - 5)
অতএব, \frac{1}{{{x^2} - 5x + 6}}

= \frac{1}{{(x - 2)(x - 3)}} \times \frac{{(x - 4)(x - 5)}}{{(x - 4)(x - 5)}}  [\because (x - 2)(x - 3)(x - 4)(x - 5) \div (x - 2)(x - 3) = (x - 4)(x - 5)]

= \frac{{(x - 4)(x - 5)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}}

এখন, \frac{1}{{{x^2} - 7x + 12}}

= \frac{1}{{(x - 3)(x - 4)}} \times \frac{{(x - 2)(x - 5)}}{{(x - 2)(x - 5)}}    [\because (x - 2)(x - 3)(x - 4)(x - 5) \div (x - 3)(x - 4) = (x - 2)(x - 5)]

= \frac{{(x - 2)(x - 5)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}}

এবং \frac{1}{{{x^2} - 9x + 20}}

= \frac{1}{{(x - 4)(x - 5)}} \times \frac{{(x - 2)(x - 3)}}{{(x - 2)(x - 3)}}  [\because (x - 2)(x - 3)(x - 4)(x - 5) \div (x - 4)(x - 5) = (x - 2)(x - 3)]

= \frac{{(x - 2)(x - 3)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}}

সাধারন হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
\frac{{(x - 4)(x - 5)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}}, \frac{{(x - 2)(x - 5)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}}, \frac{{(x - 2)(x - 3)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}}    (Ans.)

সমাধান ২(ছ):
\frac{{a - b}}{{{a^2}{b^2}}},\frac{{b - c}}{{{b^2}{c^2}}},\frac{{c - a}}{{{c^2}{a^2}}}
এখানে,
১ম ভগ্নাংশের হর ={{a^2}{b^2}}
২য় ভগ্নাংশের হর { = {b^2}{c^2}}
৩য় ভগ্নাংশের হর { = {c^2}{a^2}}
\therefore হরগুলোর ল.সা.গু {a^2}{b^2}{c^2}
অতএব, \frac{{a - b}}{{{a^2}{b^2}}}
= \frac{{a - b}}{{{a^2}{b^2}}} \times \frac{{{c^2}}}{{{c^2}}}   [\because {a^2}{b^2}{c^2} \div {a^2}{b^2} = {c^2}]
= \frac{{(a - b){c^2}}}{{{a^2}{b^2}{c^2}}}

এখন, \frac{{b - c}}{{{b^2}{c^2}}}

= \frac{{b - c}}{{{b^2}{c^2}}} \times \frac{{{a^2}}}{{{a^2}}}    [\because {a^2}{b^2}{c^2} \div {b^2}{c^2} = {a^2}]

= \frac{{(b - c){a^2}}}{{{a^2}{b^2}{c^2}}}

এবং \frac{{c - a}}{{{c^2}{a^2}}}

= \frac{{c - a}}{{{c^2}{a^2}}} \times \frac{{{b^2}}}{{{b^2}}}    [\because {a^2}{b^2}{c^2} \div {c^2}{a^2} = {b^2}]

= \frac{{(c - a){b^2}}}{{{a^2}{b^2}{c^2}}}

সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
\frac{{(a - b){c^2}}}{{{a^2}{b^2}{c^2}}}, \frac{{(b - c){a^2}}}{{{a^2}{b^2}{c^2}}}, \frac{{(c - a){b^2}}}{{{a^2}{b^2}{c^2}}} (Ans.)

সমাধান ২(জ):
\frac{{x - y}}{{x + y}},\frac{{y - z}}{{y + z}},\frac{{z - x}}{{z + x}}
১ম ভগ্নাংশের হর { = x + y}
২য় ভগ্নাংশের হর { = y + z}
৩য় ভগ্নাংশের হর { = z + x}
\therefore হরগুলোর ল.সা.গু = (x + y)(y + z)(z + x)

অতএব, \frac{{x - y}}{{x + y}}

= \frac{{x - y}}{{x + y}} \times \frac{{(y + z)(z + x)}}{{(y + z)(z + x)}}   [\because (x + y)(y + z)(z + x) \div (x + y) = (y + z)(z + x)]

= \frac{{(x - y)(y + z)(z + x)}}{{(x + y)(y + z)(z + x)}}

এখন, \frac{{y - z}}{{y + z}}

= \frac{{y - z}}{{y + z}} \times \frac{{(x + y)(z + x)}}{{(x + y)(z + x)}}   [\because (x + y)(y + z)(z + x) \div (y + z) = (x + y)(z + x)]

= \frac{{(y - z)(x + y)(z + x)}}{{(x + y)(y + z)(z + x)}}

এবং \frac{{z - x}}{{z + x}}

= \frac{{z - x}}{{z + x}} \times \frac{{(x + y)(y + z)}}{{(x + y)(y + z)}}   [\because (x + y)(y + z)(z + x) \div (z + x) = (x + y)(y + z)]

= \frac{{(z - x)(x + y)(y + z)}}{{(x + y)(y + z)(z + x)}}

\therefore সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো নিম্নরূপ:
\frac{{(x - y)(y + z)(z + x)}}{{(x + y)(y + z)(z + x)}}, \frac{{(y - z)(x + y)(z + x)}}{{(x + y)(y + z)(z + x)}}, \frac{{(z - x)(x + y)(y + z)}}{{(x + y)(y + z)(z + x)}}    (Ans.)

৩। যোগফল নির্ণয় কর:
(ক)\frac{{a - b}}{a} + \frac{{a + b}}{b}       (খ)\frac{a}{{bc}} + \frac{b}{{ca}} + \frac{c}{{ab}}

(গ)\frac{{x - y}}{x} + \frac{{y - z}}{y} + \frac{{z - x}}{z}    (ঘ)\frac{{x + y}}{{x - y}} + \frac{{x - y}}{{x + y}}

(ঙ)\frac{1}{{{x^2} - 3x + 2}} + \frac{1}{{{x^2} - 4x + 3}} + \frac{1}{{{x^2} - 5x + 4}}

(চ)\frac{1}{{{a^2} - {b^2}}} + \frac{1}{{{a^2} + ab + {b^2}}} + \frac{1}{{{a^2} - ab + {b^2}}}

(ছ)\frac{1}{{x - 2}} + \frac{1}{{x + 2}} + \frac{4}{{{x^2} - 4}}

(জ)\frac{1}{{{x^2} - 1}} + \frac{1}{{{x^4} - 1}} + \frac{4}{{{x^8} - 1}}

সমাধান ৩(ক):
প্রদত্ত রাশি = \frac{{a - b}}{a} + \frac{{a + b}}{b}

= \frac{{b(a - b) + a(a + b)}}{{ab}}

= \frac{{ab - {b^2} + {a^2} + ab}}{{ab}}

= \frac{{{a^2} + 2ab - {b^2}}}{{ab}}     (Ans.)

সমাধান ৩(খ):
প্রদত্ত রাশি = \frac{a}{{bc}} + \frac{b}{{ca}} + \frac{c}{{ab}}

= \frac{{a.a + b.b + c.c}}{{abc}}

= \frac{{{a^2} + {b^2} + {c^2}}}{{abc}}     (Ans.)

সমাধান ৩(গ):
প্রদত্ত রাশি= \frac{{x - y}}{x} + \frac{{y - z}}{y} + \frac{{z - x}}{z}

= \frac{{yz(x - y) + zx(y - z) + xy(z - x)}}{{xyz}}

= \frac{{xyz - {y^2}z + xyz - {z^2}x + xyz - {x^2}y}}{{xyz}}

= \frac{{3xyz - {x^2}y - {y^2}z - {z^2}x}}{{xyz}}    (Ans.)

সমাধান ৩(ঘ):
প্রদত্ত রাশি= \frac{{x + y}}{{x - y}} + \frac{{x - y}}{{x + y}}

= \frac{{{{(x + y)}^2} + {{(x - y)}^2}}}{{(x - y)(x + y)}}

= \frac{{{x^2} + 2xy + {y^2} + {x^2} - 2xy + {y^2}}}{{{x^2} - {y^2}}}

= \frac{{2{x^2} + 2{y^2}}}{{{x^2} - {y^2}}}

= \frac{{2({x^2} + {y^2})}}{{{x^2} - {y^2}}}     (Ans.)

সমাধান ৩(ঙ):
প্রদত্ত রাশি= \frac{1}{{{x^2} - 3x + 2}} + \frac{1}{{{x^2} - 4x + 3}} + \frac{1}{{{x^2} - 5x + 4}}

= \frac{1}{{{x^2} - 2x - x + 2}} + \frac{1}{{{x^2} - 3x - x + 3}} + \frac{1}{{{x^2} - 4x - x + 4}}

= \frac{1}{{x(x - 2) - 1(x - 2)}} + \frac{1}{{x(x - 3) - 1(x - 3)}} + \frac{1}{{x(x - 4) - 1(x - 4)}}

= \frac{1}{{(x - 2)(x - 1)}} + \frac{1}{{(x - 3)(x - 1)}} + \frac{1}{{(x - 4)(x - 1)}}

= \frac{{(x - 3)(x - 4) + (x - 2)(x - 4) + (x - 2)(x - 3)}}{{(x - 1)(x - 2)(x - 3)(x - 4)}}

= \frac{{{x^2} - 4x - 3x + 12 + {x^2} - 2x - 4x + 8 + {x^2} - 2x - 3x + 6}}{{(x - 1)(x - 2)(x - 3)(x - 4)}}

= \frac{{3{x^2} - 18x + 26}}{{(x - 1)(x - 2)(x - 3)(x - 4)}}      (Ans.)

সমাধান ৩(চ):
প্রদত্ত রাশি= \frac{1}{{{a^2} - {b^2}}} + \frac{1}{{{a^2} + ab + {b^2}}} + \frac{1}{{{a^2} - ab + {b^2}}}

= \frac{{\left( {{a^2} + ab + {b^2}} \right)\left( {{a^2} - ab + {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\left( {{a^2} - ab + {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\left( {{a^2} + ab + {b^2}} \right)}}{{\left( {{a^2} - {b^2}} \right)\left( {{a^2} + ab + {b^2}} \right)\left( {{a^2} - ab + {b^2}} \right)}}

= \frac{{\left( {{a^2} + {b^2} + ab} \right)\left( {{a^2} + {b^2} - ab} \right) + (a + b)(a - b)({a^2} - ab + {b^2}) + (a + b)(a - b)({a^2} + ab + {b^2})}}{{(a + b)(a - b)({a^2} + ab + {b^2})({a^2} - ab + {b^2})}}

= \frac{{{{({a^2} + {b^2})}^2} - {{(ab)}^2} + (a - b)(a + b)({a^2} - ab + {b^2}) + (a + b)(a - b)({a^2} + ab + {b^2})}}{{(a + b)({a^2} - ab + {b^2})(a - b)({a^2} + ab + {b^2})}}

= \frac{{{{({a^2})}^2} + 2{a^2}{b^2} + {{({b^2})}^2} - {a^2}{b^2} + (a - b)({a^3} + {b^3}) + (a + b)({a^3} - {b^3})}}{{({a^3} + {b^3})({a^3} - {b^3})}}

= \frac{{{a^4} + 2{a^2}{b^2} + {b^4} - {a^2}{b^2} + {a^4} + a{b^3} - {a^3}b - {b^4} + {a^4} - a{b^3} + {a^3}b - {b^4}}}{{({a^3} + {b^3})({a^3} - {b^3})}}

= \frac{{({a^4} + {a^4} + {a^4}) + (2{a^2}{b^2} - {a^2}{b^2}) + (a{b^3} - {a^3}b - a{b^3}\_{a^3}b) + ({b^4} - {b^4} - {b^4})}}{{({a^3} + {b^3})({a^3} - {b^3})}}

= \frac{{3{a^4} + {a^2}{b^2} - {b^4}}}{{({a^3} + {b^3})({a^3} - {b^3})}}

সমাধান ৩(ছ):
প্রদত্ত রাশি= \frac{1}{{x - 2}} + \frac{1}{{x + 2}} + \frac{4}{{{x^2} - 4}}

= \frac{1}{{x - 2}} + \frac{1}{{x + 2}} + \frac{4}{{{x^2} - {2^2}}}

= \frac{1}{{x - 2}} + \frac{1}{{x + 2}} + \frac{4}{{(x + 2)(x - 2)}}

= \frac{{x + 2 + x - 2 + 4}}{{(x + 2)(x - 2)}}

= \frac{{2x + 4}}{{(x + 2)(x - 2)}}

= \frac{{2(x + 2)}}{{(x + 2)(x - 2)}}

= \frac{2}{{x - 2}}     (Ans.)

সমাধান ৩(জ):
প্রদত্ত রাশি= \frac{1}{{{x^2} - 1}} + \frac{1}{{{x^4} - 1}} + \frac{4}{{{x^8} - 1}}

= \frac{1}{{{x^2} - 1}} + \frac{1}{{{{({x^2})}^2} - 1}} + \frac{4}{{{{({x^4})}^2} - 1}}

= \frac{1}{{{x^2} - 1}} + \frac{1}{{({x^2} + 1)({x^2} - 1)}} + \frac{4}{{({x^4} + 1)({x^4} - 1)}}

= \frac{1}{{{x^2} - 1}} + \frac{1}{{({x^2} + 1)({x^2} - 1)}} + \frac{4}{{({x^4} + 1)({x^2} + 1)({x^2} - 1)}}

= \frac{{({x^4} + 1)({x^2} + 1) + {x^4} + 1 + 4}}{{({x^4} + 1)({x^2} + 1)({x^2} - 1)}}

= \frac{{{x^4}.{x^2} + {x^4}.1 + 1.{x^2} + 1.1 + {x^4} + 1 + 4}}{{({x^4} + 1)\{ {{({x^2})}^2} - {{(1)}^2}\} }}

= \frac{{{x^6} + {x^4} + {x^2} + 1 + {x^4} + 5}}{{({x^4} + 1)({x^4} - 1)}}

= \frac{{{x^6} + 2{x^4} + {x^2} + 6}}{{{{({x^4})}^2} - {{(1)}^2}}}

= \frac{{{x^6} + 2{x^4} + {x^2} + 6}}{{{x^8} - 1}}

৪। বিয়োগফল নির্ণয় কর: (ক) \frac{a}{{x - 3}} - \frac{{{a^2}}}{{{x^2} - 9}}       (খ) \frac{1}{{y(x - y)}} - \frac{1}{{x(x + y)}}

(গ) \frac{{x + 1}}{{1 + x + {x^2}}} - \frac{{x - 1}}{{1 - x + {x^2}}} (ঘ) \frac{{{a^2} + 16{b^2}}}{{{a^2} - 16{b^2}}} - \frac{{a - 4b}}{{a + 4b}}

(ঙ) \frac{1}{{x - y}} - \frac{{{x^2} - xy + {y^2}}}{{{x^3} + {y^3}}}

সমাধান ৪(ক):
প্রদত্ত রাশি = \frac{a}{{x - 3}} - \frac{{{a^2}}}{{{x^2} - 9}}

= \frac{a}{{x - 3}} - \frac{{{a^2}}}{{{x^2} - {3^2}}}

= \frac{a}{{x - 3}} - \frac{{{a^2}}}{{(x + 3)(x - 3)}}    [\because {a^2} - {b^2} = (a + b)(a - b)]

= \frac{{a(x + 3) - {a^2}}}{{(x + 3)(x - 3)}}

= \frac{{ax + 3a - {a^2}}}{{{x^2} - 9}}     (Ans.)

সমাধান ৪(খ):
প্রদত্ত রাশি= \frac{1}{{y(x - y)}} - \frac{1}{{x(x + y)}}

= \frac{{x(x + y) - y(x - y)}}{{xy(x + y)(x - y)}}

= \frac{{{x^2} + xy - xy + {y^2}}}{{xy({x^2} - {y^2})}}    [\because {a^2} - {b^2} = (a + b)(a - b)]

= \frac{{{x^2} + {y^2}}}{{xy({x^2} - {y^2})}}    (Ans.)

সমাধান ৪(গ):
প্রদত্ত রাশি = \frac{{x + 1}}{{1 + x + {x^2}}} - \frac{{x - 1}}{{1 - x + {x^2}}}

= \frac{{\left( {x + 1} \right)\left( {1 - x + {x^2}} \right) - \left( {x - 1} \right)\left( {1 + x + {x^2}} \right)}}{{\left( {1 + x + {x^2}} \right)\left( {1 - x + {x^2}} \right)}}

= \frac{{\left( {{x^3} + {1^3}} \right) - \left( {{x^3} - {1^3}} \right)}}{{\{ (1 + {x^2}) + x\} \{ (1 + {x^2}) - x\} }}

= \frac{{{x^3} + 1 - {x^3} + 1}}{{{{(1 + {x^2})}^2} - {x^2}}}    [\because {a^2} - {b^2} = (a + b)(a - b)]

= \frac{2}{{1 + 2.1.{x^2} + {{({x^2})}^2} - {x^2}}}    [\because {(a + b)^2} = {a^2} + 2ab + {b^2}]

= \frac{2}{{1 + 2{x^2} + {x^4} - {x^2}}}

= \frac{2}{{1 + {x^2} + {x^4}}}

= \frac{2}{{{x^4} + {x^2} + 1}}

সমাধান ৪(ঘ):
প্রদত্ত রাশি= \frac{{{a^2} + 16{b^2}}}{{{a^2} - 16{b^2}}} - \frac{{a - 4b}}{{a + 4b}}

= \frac{{{a^2} + 16{b^2}}}{{{a^2} - {{(4b)}^2}}} - \frac{{a - 4b}}{{a + 4b}}

= \frac{{{a^2} + 16{b^2}}}{{(a + 4b)(a - 4b)}} - \frac{{a - 4b}}{{a + 4b}}   [\because {a^2} - {b^2} = (a + b)(a - b)]

= \frac{{{a^2} + 16{b^2} - {{(a - 4b)}^2}}}{{(a + 4b)(a - 4b)}}

= \frac{{{a^2} + 16{b^2} - \{ {a^2} - 2.a.4b + {{(4b)}^2}\} }}{{{a^2} - 16{b^2}}}   [\because {(a - b)^2} = {a^2} - 2ab + {b^2}]

= \frac{{{a^2} + 16{b^2} - ({a^2} - 8ab + 16{b^2})}}{{{a^2} - 16{b^2}}}

= \frac{{{a^2} + 16{b^2} - {a^2} + 8ab - 16{b^2}}}{{{a^2} - 16{b^2}}}

= \frac{{8ab}}{{{a^2} - 16{b^2}}}

সমাধান ৪(ঙ):
প্রদত্ত রাশি= \frac{1}{{x - y}} - \frac{{{x^2} - xy + {y^2}}}{{{x^3} + {y^3}}}

= \frac{1}{{x - y}} - \frac{{{x^2} - xy + {y^2}}}{{(x + y)({x^2} - xy + {y^2})}}   [যেহেতু{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})]

= \frac{1}{{x - y}} - \frac{1}{{x + y}}

= \frac{{x + y - (x - y)}}{{(x + y)(x - y)}}

= \frac{{x + y - x + y}}{{{x^2} - {y^2}}}

= \frac{{2y}}{{{x^2} - {y^2}}}     (Ans.)

৫। সরল কর:
(ক) \frac{{x - y}}{{xy}} + \frac{{y - z}}{{yz}} + \frac{{z - x}}{{zx}}

(খ) \frac{{x - y}}{{(x + y)(y + z)}} + \frac{{y - z}}{{(y + z)(z + x)}} + \frac{{z - x}}{{(z + x)(x + y)}}

(গ) \frac{y}{{(x - y)(y - z)}} + \frac{x}{{(z - x)(x - y)}} + \frac{z}{{(y - z)(z - x)}}

(ঘ)\frac{1}{{x + 3y}} + \frac{1}{{x - 3y}} - \frac{{2x}}{{{x^2} - 9{y^2}}}

(ঙ) \frac{1}{{x - y}} - \frac{2}{{2x + y}} + \frac{1}{{x + y}} - \frac{2}{{2x - y}}

(চ) \frac{1}{{x - 2}} - \frac{{x - 2}}{{{x^2} + 2x + 4}} + \frac{{6x}}{{{x^3} + 8}}

(ছ) \frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} + \frac{4}{{{x^4} + 1}}

(জ) \frac{{x - y}}{{(y - z)(z - x)}} + \frac{{y - z}}{{(z - x)(x - y)}} + \frac{{z - x}}{{(x - y)(y - z)}}

(ঝ) \frac{1}{{a - b - c}} + \frac{1}{{a - b + c}} + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

(ঞ) \frac{1}{{{a^2} + {b^2} - {c^2} + 2ab}} + \frac{1}{{{b^2} + {c^2} - {a^2} + 2bc}} + \frac{1}{{{c^2} + {a^2} - {b^2} + 2ca}}

সমাধান ৫(ক):
প্রদত্ত রাশি = \frac{{x - y}}{{xy}} + \frac{{y - z}}{{yz}} + \frac{{z - x}}{{zx}}

= \frac{{z(x - y) + x(y - z) + y(z - x)}}{{xyz}}

= \frac{{zx - zy + xy - zx + zy - xy}}{{xyz}}

= \frac{0}{{xyz}}

= 0

সমাধান ৫(খ):
প্রদত্ত রাশি= \frac{{x - y}}{{(x + y)(y + z)}} + \frac{{y - z}}{{(y + z)(z + x)}} + \frac{{z - x}}{{(z + x)(x + y)}}

= \frac{{(x - y)(z + x) + (y - z)(x + y) + (z - x)(y + z)}}{{(x + y)(y + z)(z + x)}}

= \frac{{xz + {x^2} - yz - xy + yx + {y^2} - zx - zy + zy + {z^2} - xy - xz}}{{(x + y)(y + z)(z + x)}}
= \frac{{{x^2} + {y^2} + {z^2} - xy - yz - zx}}{{(x + y)(y + z)(z + x)}}

সমাধান ৫(গ):
প্রদত্ত রাশি= \frac{y}{{(x - y)(y - z)}} + \frac{x}{{(z - x)(x - y)}} + \frac{z}{{(y - z)(z - x)}}
= \frac{{y(z - x) + x(y - z) + z(x - y)}}{{(x - y)(y - z)(z - x)}}

= \frac{{yz - yx + xy - xz + zx - zy}}{{(x - y)(y - z)(z - x)}}

= \frac{0}{{(x - y)(y - z)(z - x)}}

= 0    (Ans.)

সমাধান ৫(ঘ):
প্রদত্ত রাশি= \frac{1}{{x + 3y}} + \frac{1}{{x - 3y}} - \frac{{2x}}{{{x^2} - 9{y^2}}}

= \frac{1}{{x + 3y}} + \frac{1}{{x - 3y}} - \frac{{2x}}{{{x^2} - {{(3y)}^2}}}

= \frac{1}{{x + 3y}} + \frac{1}{{x - 3y}} - \frac{{2x}}{{(x + 3y)(x - 3y)}}

= \frac{{x - 3y + x + 3y - 2x}}{{(x + 3y)(x - 3y)}}

= \frac{{2x - 3y + 3y - 2x}}{{(x + 3y)(x - 3y)}}

= \frac{0}{{{x^2} - 9{y^2}}}

= 0      (Ans.)

সমাধান ৫(ঙ):
প্রদত্ত রাশি= \frac{1}{{x - y}} - \frac{2}{{2x + y}} + \frac{1}{{x + y}} - \frac{2}{{2x - y}}

= \frac{1}{{x - y}} + \frac{1}{{x + y}} - \frac{2}{{2x + y}} - \frac{2}{{2x - y}}

= \frac{{x + y + x - y}}{{(x - y)(x + y)}} - \left( {\frac{2}{{2x + y}} + \frac{2}{{2x - y}}} \right)

= \frac{{2x}}{{(x + y)(x - y)}} - \left\{ {\frac{{2(2x - y) + 2(2x + y)}}{{(2x + y)(2x - y)}}} \right\}

= \frac{{2x}}{{{x^2} - {y^2}}} - \left\{ {\frac{{4x - 2y + 4x + 2y}}{{{{(2x)}^2} - {y^2}}}} \right\}

= \frac{{2x}}{{{x^2} - {y^2}}} - \frac{{8x}}{{4{x^2} - {y^2}}}

= \frac{{2x(4{x^2} - {y^2}) - 8x({x^2} - {y^2})}}{{({x^2} - {y^2})(4{x^2} - {y^2})}}

= \frac{{8{x^3} - 2x{y^2} - 8{x^3} + 8x{y^2}}}{{({x^2} - {y^2})(4{x^2} - {y^2})}}

= \frac{{6x{y^2}}}{{({x^2} - {y^2})(4{x^2} - {y^2})}}

সমাধান (চ):
প্রদত্ত রাশি= \frac{1}{{x - 2}} - \frac{{x - 2}}{{{x^2} + 2x + 4}} + \frac{{6x}}{{{x^3} + 8}}

= \left( {\frac{1}{{x - 2}} - \frac{{x - 2}}{{{x^2} + 2x + 4}}} \right) + \frac{{6x}}{{{x^3} + 8}}

= \frac{{{x^2} + 2x + 4 - {{(x - 2)}^2}}}{{(x - 2)({x^2} + 2x + 4)}} + \frac{{6x}}{{{x^3} + 8}}

= \frac{{{x^2} + 2x + 4 - ({x^2} - 2x.2 + {2^2})}}{{{x^3} - {2^3}}} + \frac{{6x}}{{{x^3} + 8}}

= \frac{{{x^2} + 2x + 4 - {x^2} + 4x - 4}}{{{x^3} - 8}} + \frac{{6x}}{{{x^3} + 8}}

= \frac{{6x}}{{{x^3} - 8}} + \frac{{6x}}{{{x^3} + 8}}

= \frac{{6x({x^3} + 8) + 6x({x^3} - 8)}}{{({x^3} - 8)({x^3} + 8)}}

= \frac{{6{x^4} + 48x + 6{x^4} - 48x}}{{{{({x^3})}^2} - {8^2}}}

= \frac{{12{x^4}}}{{{x^6} - 64}}  

সমাধান (ছ):
প্রদত্ত রাশি= \frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} + \frac{4}{{{x^4} + 1}}

= \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) - \frac{2}{{{x^2} + 1}} + \frac{4}{{{x^4} + 1}}

= \frac{{x + 1 - (x - 1)}}{{(x + 1)(x - 1)}} - \frac{2}{{{x^2} + 1}} + \frac{4}{{{x^4} + 1}}

= \frac{{x + 1 - x + 1}}{{{x^2} - 1}} - \frac{2}{{{x^2} + 1}} + \frac{4}{{{x^4} + 1}}

= \frac{2}{{{x^2} - 1}} - \frac{2}{{{x^2} + 1}} + \frac{4}{{{x^4} + 1}}

= \frac{{2({x^2} + 1) - 2({x^2} - 1)}}{{({x^2} + 1)({x^2} - 1)}} + \frac{4}{{{x^4} + 1}}

= \frac{{2{x^2} + 2 - 2{x^2} + 2}}{{{{({x^2})}^2} - {1^2}}} + \frac{4}{{{x^4} + 1}}

= \frac{4}{{{x^4} - 1}} + \frac{4}{{{x^4} + 1}}

= \frac{{4({x^4} + 1) + 4({x^4} - 1)}}{{({x^4} + 1)({x^4} - 1)}}

= \frac{{4{x^4} + 4 + 4{x^4} - 4}}{{{{({x^4})}^2} - {1^2}}}

= \frac{{8{x^4}}}{{{x^8} - 1}}

সমাধান (জ):
প্রদত্ত রাশি= \frac{{x - y}}{{(y - z)(z - x)}} + \frac{{y - z}}{{(z - x)(x - y)}} + \frac{{z - x}}{{(x - y)(y - z)}}

= \frac{{(x - y)(x - y) + (y - z)(y - z) + (z - x)(z - x)}}{{(x - y)(y - z)(z - x)}}

= \frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{(x - y)(y - z)(z - x)}}

= \frac{{{x^2} - 2xy + {y^2} + {y^2} - 2yz + {z^2} + {z^2} - 2zx + {x^2}}}{{(x - y)(y - z)(z - x)}}

= \frac{{2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx}}{{(x - y)(y - z)(z - x)}}

= \frac{{2({x^2} + {y^2} + {z^2} - xy - yz - zx)}}{{(x - y)(y - z)(z - x)}}

সমাধান (ঝ):
প্রদত্ত রাশি= \frac{1}{{a - b - c}} + \frac{1}{{a - b + c}} + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \left( {\frac{1}{{a - b - c}} + \frac{1}{{a - b + c}}} \right) + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \frac{{a - b + c + a - b - c}}{{(a - b - c)(a - b + c)}} + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \frac{{2a - 2b}}{{\{ (a - b) - c\} \{ (a - b) + c\} }} + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \frac{{2a - 2b}}{{{{(a - b)}^2} - {c^2}}} + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \frac{{2a - 2b}}{{{a^2} - 2ab + {b^2} - {c^2}}} + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \frac{{2a - 2b}}{{{a^2} + {b^2} - {c^2} - 2ab}} + \frac{a}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \frac{{2a - 2b + a}}{{{a^2} + {b^2} - {c^2} - 2ab}}

= \frac{{3a - 2b}}{{{a^2} + {b^2} - {c^2} - 2ab}}

সমাধান (ঞ):
প্রদত্ত রাশি= \frac{1}{{{a^2} + {b^2} - {c^2} + 2ab}} + \frac{1}{{{b^2} + {c^2} - {a^2} + 2bc}} + \frac{1}{{{c^2} + {a^2} - {b^2} + 2ca}}

= \frac{1}{{({a^2} + 2ab + {b^2}) - {c^2}}} + \frac{1}{{({b^2} + 2bc + {c^2}) - {a^2}}} + \frac{1}{{({c^2} + 2ca + {a^2}) - {b^2}}}

= \frac{1}{{{{(a + b)}^2} - {c^2}}} + \frac{1}{{{{(b + c)}^2} - {a^2}}} + \frac{1}{{{{(c + a)}^2} - {b^2}}}

= \frac{1}{{(a + b + c)(a + b - c)}} + \frac{1}{{(b + c + a)(b + c - a)}} + \frac{1}{{(c + a + b)(c + a - b)}}

= \frac{{b + c - a + a + b - c}}{{(a + b + c)(a + b - c)(b + c - a)}} + \frac{1}{{(c + a + b)(c + a - b)}}

= \frac{{2b}}{{(a + b + c)(a + b - c)(b + c - a)}} + \frac{1}{{(c + a + b)(c + a - b)}}

= \frac{{2b(c + a - b) + (a + b - c)(b + c - a)}}{{(a + b + c)(a + b - c)(b + c - a)(c + a - b)}}

= \frac{{2bc + 2ab - 2{b^2} + ab + ac - {a^2} + {b^2} + bc - ab - bc - {c^2} + ca}}{{(a + b + c)(a + b - c)(b + c - a)(c + a - b)}}

= \frac{{2ab + 2bc + ac + ca + ab - ab + bc - bc - {a^2} - 2{b^2} + {b^2} - {c^2}}}{{(a + b + c)(a + b - c)(b + c - a)(c + a - b)}}

= \frac{{2ab + 2bc + 2ca - {a^2} - {b^2} - {c^2}}}{{(a + b + c)(a + b - c)(b + c - a)(c + a - b)}}

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