If alpha and beta are the roots of the equati | vedclass

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(B) Given $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$.Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:$= 1 + (\alpha + \b

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(B) Given $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$.Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:$= 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha \beta + \alpha \beta (\alpha + \beta) + \alpha^2 \beta^2$$= 1 + (\alpha + \beta) + ((\alpha + \beta)^2 - 2\alpha \beta) + \alpha \beta + \alpha \beta (\alpha + \beta) + (\alpha \beta)^2$$= 1 + (\alpha + \beta) + (\alpha + \beta)^2 - \alpha \beta + \alpha \beta (\alpha + \beta) + (\alpha \beta)^2$Substituting the values:$= 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{c}{a} + \left(\frac{c}{a}\right)\left(-\frac{b}{a}\right) + \frac{c^2}{a^2}$$= \frac{a^2 - ab + b^2 - ac - bc + c^2}{a^2} = \frac{a^2 + b^2 + c^2 - ab - bc - ca}{a^2}$$= \frac{1}{2a^2} [(a - b)^2 + (b - c)^2 + (c - a)^2]$Since $a, b, c$ are distinct,the sum of squares is always positive. Thus,the expression is always positive.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ ($a \ne 0$; $a, b, c$ being distinct),then $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2) = $

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