How can one solve this without expanding. \begin{array}{l}
\text { If } a+b+c=1, a b+b c+c a=2 \\
\text { and } a b c=3 \text {. What is the value } \\
\text { of } a^{4}+b^{4}+c^{4} \text { ? }
\end{array}
This can be solved by expanding but is there any easy alternative method ?
Here is how I solved :
\begin{array}{l}
(a+b+c)^{2}=1\\
\Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=1\\
\Rightarrow a^{2}+b^{2}+c^{2}=-3 \ldots . .(i)\\
\Rightarrow a b+b c+c a=2 \ldots . \text { (ii) }\\
\text { Squaring of equation (ii), }\\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2\left(a b^{2} c+b c^{2} a+\right.\\
\left.c a^{2} b\right)=4\\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)=4\\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+6=4
\end{array}
\begin{array}{l}
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=-2 \ldots . \text { (iii) }\\
\text { Squaring of equation (i), }\\
\Rightarrow a^{4}+b^{4}+c^{4}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)=9\\
\Rightarrow a^{4}+b^{4}+c^{4}-4=9\\
\Rightarrow a^{4}+b^{4}+c^{4}=13
\end{array}

$a,b,c$ are roots of $x^3 - x^2 + 2x-3 = 0$
$$a^2+b^2+c^2=(a+b+c)^2 - 2(ab+bc+ac) = 1-2(2)=-3$$
Since
$$x^3 = x^2 -2x+3$$
$$a^3+b^3+c^3 = (a^2+b^2+c^2) - 2(a+b+c) + 9 = -3-2(1)+9=4$$
Since
$$x^4 = x^3 -2x^2+3x$$
$$a^4+b^4+c^4 = 4-2(-3)+3(1)=13$$
