Question

if a+b+c=0 then prove that (b+c/3bc)square + (c+a/3bc)square + (a+b/3ab)square=1

Answer 1

$\huge\mathfrak\red{\underline{\underline{Solution :-}}}$

$\textbf{L.H.S=}$ $\huge\sf\frac{(b+c)^2}{3bc}$ + $\huge\sf\frac{(c+a)^2}{3ca}$+$\huge\sf\frac{(a+b)^2}{3ab}$

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$\huge\sf\frac{(-a)^2}{3bc}$+$\huge\sf\frac{(-b)^2}{3ca}$+$\huge\sf\frac{(-c)^2}{3ab}$

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$\huge\sf\frac{a^2}{3bc}$+ $\huge\sf\frac{b^2}{3ca}$+ $\huge\sf\frac{c^2}{3ab}$

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$\huge\sf\frac{a^3+b^3+c^3}{3abc}$

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$\huge\sf\frac{3abc}{3abc}$

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$\huge\pink{\boxed{\boxed{Ans.=1}}}$

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= R.H.S