If $a + b + c = 0$ , prove that identities $2 (a^{4} + b^{4} + c^{4}) = (a^{2} + b^{2} + c^{2})^{2}$ .

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Solution

$2 (a^{4} + b^{4} + c^{4}) = 2 ((a^{2} + b^{2} + c^{2})^{2} - 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}))$
$= 2 ((a^{2} + b^{2} + c^{2})^{2} - 2 ((a b + b c + c a)^{2} - 2 a b c (a + b + c))$
we know that $a + b + c = 0$
so, equation deduces to
$= 2 ((a^{2} + b^{2} + c^{2})^{2} - 2 (a b + b c + c a)^{2})$
$a + b + c = 0, a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 0$
$a b + b c + c a = \frac{- (a^{2} + b^{2} + c^{2})}{2}$
substitutung this value in the expression, we get
$= 2 ((a^{2} + b^{2} + c^{2})^{2} - 2 \frac{(a^{2} + b^{2} + c^{2})^{2}}{4})$
$= (a^{2} + b^{2} + c^{2})^{2}$

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