If alpha, beta, gamma are the roots of the eq | vedclass

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Question

(D) For the cubic equation $x^3 - 6x^2 + 11x - 6 = 0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\sum \alpha = 6$,
$\sum \a

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(D) For the cubic equation $x^3 - 6x^2 + 11x - 6 = 0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\sum \alpha = 6$,
$\sum \alpha \beta = 11$,
$\alpha \beta \gamma = 6$.
We need to find $\sum \alpha^2 \beta + \sum \alpha \beta^2$.
This expression can be written as $(\sum \alpha \beta)(\sum \alpha) - 3 \alpha \beta \gamma$.
Substituting the values:
$(11)(6) - 3(6) = 66 - 18 = 48$.
Wait,let us re-evaluate the expression:
$(\alpha + \beta + \gamma)(\alpha \beta + \beta \gamma + \gamma \alpha) = \sum \alpha^2 \beta + \sum \alpha \beta^2 + 3 \alpha \beta \gamma$.
So,$\sum \alpha^2 \beta + \sum \alpha \beta^2 = (\sum \alpha)(\sum \alpha \beta) - 3 \alpha \beta \gamma$.
$= (6)(11) - 3(6) = 66 - 18 = 48$.
Given the options provided,there might be a typo in the original equation constant. If the equation was $x^3 - 6x^2 + 11x - 6 = 0$,the roots are $1, 2, 3$.
Then $\sum \alpha^2 \beta + \sum \alpha \beta^2 = (1^2 \cdot 2 + 1^2 \cdot 3 + 2^2 \cdot 1 + 2^2 \cdot 3 + 3^2 \cdot 1 + 3^2 \cdot 2) = 2 + 3 + 4 + 12 + 9 + 18 = 48$.
If the constant was different,the result would change. Based on the standard form,the answer is $48$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x-6=0$,then $\sum \alpha^2 \beta + \sum \alpha \beta^2 =$

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