If $a + b + c = 1, a b + b c + c a = 2$ and $a b c = 3$ , then the value of $a^{4} + b^{4} + c^{4}$ is equal to

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Solution

$∵ a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (a b + b c + c a)$
$∴ a^{2} + b^{2} + c^{2} = 1 - 4 = - 3$
and $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = (a b + b c + c a)^{2} - 2 a b c (a + b + c)$
$a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = 4 - 6 = - 2$
Now, $a^{4} + b^{4} + c^{4} = (a^{2} + b^{2} + c^{2})^{2} - 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})$
$\Rightarrow a^{4} + b^{4} + c^{4} = 9 + 4$
$∴ a^{4} + b^{4} + c^{4} = 13$

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