The sum of the cubes of all the roots of the | vedclass

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Question

$x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$

$x=0$ is not the root of this equation so divide it by $x^{2}$

$x^{2}-3 x-2+\frac{3}{x}+\frac{1}{x^{2}}=0$

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Vedclass · Accepted Answer

$36$

Vedclass · Answer

$x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$

$x=0$ is not the root of this equation so divide it by $x^{2}$

$x^{2}-3 x-2+\frac{3}{x}+\frac{1}{x^{2}}=0$

$x^{2}+\frac{1}{x^{2}}-2+2-3\left(x-\frac{1}{x}\right)-2=0$

$\left(x-\frac{1}{x}\right)^{2}-3\left(x-\frac{1}{x}\right)=0$

$x-\frac{1}{x}=0$

$x^{2}-1=0$

$x=\pm 1$

$\alpha=1, \beta=-1 \delta=$

$\alpha^{3}+\beta^{3}+\gamma^{3}+\delta^{3}$

$1-1+(\gamma+\delta)\left((\gamma+\delta)^{2}-3 \gamma \delta\right)$

$0+3(9-3(-1))$

$+3(12)=36$

$x-\frac{1}{x}=3$

$x^{2}-3 x-1=0$

$\gamma+\delta=3$

$\gamma \delta=-1$

The sum of the cubes of all the roots of the equation $x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$ is

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