If alpha, beta, gamma are the roots of x^3+p | vedclass

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(D) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$. From the relation between roots and coefficients: $\alpha+\b

Vedclass · Accepted Answer

$(r-p)^2+(q-1)^2$

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(D) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$. From the relation between roots and coefficients: $\alpha+\beta+\gamma = -p$ $(i)$ $\alpha\beta+\beta\gamma+\gamma\alpha = q$ $(ii)$ $\alpha\beta\gamma = -r$ $(iii)$ We know that $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = |(1+i\alpha)(1+i\beta)(1+i\gamma)|^2$. Let $f(x) = x^3+px^2+qx+r = (x-\alpha)(x-\beta)(x-\gamma)$. Then $f(i) = (i-\alpha)(i-\beta)(i-\gamma) = i^3+\pi^2+qi+r = -i-p+qi+r = (r-p) + i(q-1)$. Also $f(-i) = (-i-\alpha)(-i-\beta)(-i-\gamma) = -i^3+\pi^2-qi+r = i-p-qi+r = (r-p) - i(q-1)$. Thus,$(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = (i-\alpha)(i-\beta)(i-\gamma) 	imes (-i-\alpha)(-i-\beta)(-i-\gamma) = f(i) 	imes f(-i)$. $= ((r-p) + i(q-1))((r-p) - i(q-1)) = (r-p)^2 + (q-1)^2$.

If $\alpha, \beta, \gamma$ are the roots of $x^3+p x^2+q x+r=0$,then the value of $(1+\alpha^2)(1+\beta^2)(1+\gamma^2)$ is

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