If a+b+c=1, a b+b c+c a=2 and a b c=3,

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Question:

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

Solution:

$a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2 \Sigma a b=-3$

$(a b+b c+c a)^{2}=\Sigma(a b)^{2}+2 a b c \Sigma a$

$\Rightarrow \Sigma(a b)^{2}=-2$

$a^{4}+b^{4}+c^{4}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2 \Sigma(a b)^{2}$

$=9-2(-2)=13$

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