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

Question

Professor PC Thomas Classes · Answer

(13): Given, $a+b+c=1, a b+b c+c a$\ $=2$ and $a b c=3$\ $\because \quad(a+b+c)^{2}=1$\ $\Rightarrow \quad a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=1$ $$\begin{equation*}\Rightarrow \quad a^{2}+b^{2}+c^{2}=-3 	ag{i} \end{equation*}$$ Also, $a b+b c+c a=2$ (given)\ Squaring both sides, we get\ $(a b+b c+c a)^{2}=4$\ $\Rightarrow \quad a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2\left(a b^{2} c+b c^{2} a+ight.$ $\left.c a^{2} bight)=4$\ $\Rightarrow \quad a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}$ $+2 a b c(a+b+c)=4$\ $\Rightarrow \quad a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+6=4$ $$\begin{equation*}\Rightarrow \quad a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=-2 	ag{ii} \end{equation*}$$ Now, squaring (i), we get\ $a^{4}+b^{4}+c^{4}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}ight)=9$\ $\Rightarrow \quad a^{4}+b^{4}+c^{4}-4=9$\ $\Rightarrow \quad a^{4}+b^{4}+c^{4}=13$.\