If alpha, beta, gamma are roots of x^3-5x+4=0 | vedclass

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(D) Given the equation $x^3-5x+4=0$ with roots $\alpha, \beta, \gamma$. Comparing with $x^3+px^2+qx+r=0$,we have the sum of roots $\alpha+\beta+\gamma

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(D) Given the equation $x^3-5x+4=0$ with roots $\alpha, \beta, \gamma$. Comparing with $x^3+px^2+qx+r=0$,we have the sum of roots $\alpha+\beta+\gamma = 0$. Since $\alpha+\beta+\gamma = 0$,we use the identity: if $a+b+c=0$,then $a^3+b^3+c^3 = 3abc$. Here,$\alpha^3+\beta^3+\gamma^3 = 3\alpha\beta\gamma$. From the given equation,the product of roots $\alpha\beta\gamma = -(	ext{constant term}) = -4$. Therefore,$\alpha^3+\beta^3+\gamma^3 = 3(-4) = -12$. Finally,$(\alpha^3+\beta^3+\gamma^3)^2 = (-12)^2 = 144$.

If $\alpha, \beta, \gamma$ are roots of $x^3-5x+4=0$,then $(\alpha^3+\beta^3+\gamma^3)^2$ is equal to

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