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Question

(C) Given the equation $x^3+a x^2+b x+c=0$,the sum of the roots is $\alpha+\beta+\gamma = -a$.We can rewrite the terms in the expression as follows:$\

Vedclass · Accepted Answer

$2 a^3-9 a b+27 c$

Vedclass · Answer

(C) Given the equation $x^3+a x^2+b x+c=0$,the sum of the roots is $\alpha+\beta+\gamma = -a$.We can rewrite the terms in the expression as follows:$\alpha+\beta-2 \gamma = (\alpha+\beta+\gamma) - 3 \gamma = -a - 3 \gamma$.Similarly,$\beta+\gamma-2 \alpha = -a - 3 \alpha$ and $\gamma+\alpha-2 \beta = -a - 3 \beta$.The product is $(-a-3 \alpha)(-a-3 \beta)(-a-3 \gamma) = -(a+3 \alpha)(a+3 \beta)(a+3 \gamma)$.Let $f(x) = x^3+a x^2+b x+c = (x-\alpha)(x-\beta)(x-\gamma)$.Then $f(-a/3) = (-a/3-\alpha)(-a/3-\beta)(-a/3-\gamma) = (-1/27)(a+3 \alpha)(a+3 \beta)(a+3 \gamma)$.Thus,$(a+3 \alpha)(a+3 \beta)(a+3 \gamma) = -27 f(-a/3)$.The original expression is $-(-27 f(-a/3)) = 27 f(-a/3)$.$f(-a/3) = (-a/3)^3 + a(-a/3)^2 + b(-a/3) + c = -a^3/27 + a^3/9 - ab/3 + c = (2a^3 - 9ab + 27c)/27$.Multiplying by $27$,we get $2a^3 - 9ab + 27c$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+a x^2+b x+c=0$,then $(\alpha+\beta-2 \gamma)(\beta+\gamma-2 \alpha)(\gamma+\alpha-2 \beta)=$

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