The sum of the cubes of all the roots of the equation $x^{4}-$ $3 x^{3}-2 x^{2}+3 x+1=0$ is

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Professor PC Thomas Classes · Answer

(36) : Given $x^{4}-3 x^{3}-2 x^{2}+3 x+1=0$\ $\left(x^{2}-1ight)\left(x^{2}-3 x-1ight)=0 ; x^{2}=1$\ $\Rightarrow \quad x= \pm 1$. Now $x^{2}-3 x-1=0$\ Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x-1=0$ $\alpha+\beta=3$ and $\alpha \beta=-1$.\ Now, sum of the cubes of all roots\ $=(-1)^{3}+(1)^{3}+\alpha^{3}+\beta^{3}$\ $=-1+1+\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$\ $=27-3(-1) 	imes 3=36$\