Question

please help
I need it urgently.​

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{SOLUTION:-}}}}}}$.

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$\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$.

• $\frac{a}{(b+c)}\:+\frac{b}{(c+a)}\:+\frac{c}{(a+b)}\:=\:1$..(1)

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$\large{\underline{\underline{\mathfrak{\sf{Find\:Here:-}}}}}$.

• $\frac{a^2}{(b+c)}\:+\frac{b^2}{(c+a)}\:+\frac{c^2}{(a+b)}\:=\:?$

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$\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$.

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

Multiply both side by (a+b+c)

We get,

$\implies\frac{a}{(b+c)}\:+\frac{b}{(c+a)}\:+\frac{c}{(a+b)(a+b+c)}\:=\:(a+b+c)$

$\implies\frac{a^2}{(b+c)}\:+\frac{b^2}{(c+a)}\:+\frac{c^2}{(c+a)}\:+\frac{a\cancel{(b+c)}}{\cancel{(b+c)}}\:+\frac{b\cancel{(c+a)}}{\cancel{(c+a)}}\:+\frac{c\cancel{(a+b)}}{\cancel{(a+b)}}\:=\:(a+b+c)$

$\implies\frac{a^2}{(b+c)}\:+\frac{b^2}{(c+a)}\:+\frac{c^2}{(a+b)}\:+\:(a+b+c)\:=\:(a+b+c)$

$\implies\frac{a^2}{(b+c)}\:+\frac{b^2}{(c+a)}\:+\frac{c^2}{(a+b)}\:+\:=\:\cancel{(a+b+c)}\:-\:\cancel{(a+b+c)}$

$\bold{\boxed{\boxed{\frac{a^2}{(b+c)}\:+\frac{b^2}{(c+a)}\:+\frac{c^2}{(a+b)}\:+\:=\:0}}}$

that's proved

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