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

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

Solution

Given that $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$
$\begin{aligned} & \text { Now, }\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right) \\ & \qquad=\left(1+\alpha^2\right)\left(1+\beta^2+\gamma^2+(\beta \gamma)^2\right) \\ & =1+\beta^2+\gamma^2+(\beta \gamma)^2+\alpha^2+(\alpha \beta)^2+(\alpha \gamma)^2+(\alpha \beta \gamma)^2 \\ & =1+\left(\alpha^2+\beta^2+\gamma^2\right)+\left((\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2\right. \\ & =1+\left[(\alpha \beta \gamma)^2\right. \\ & +\left[(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)\right]+(\alpha \beta \gamma(\alpha+\beta+\gamma)]+(\alpha \beta \gamma)^2\end{aligned}$ From Eq. $(i),(i i)$ and (iii), we get $ \begin{aligned} & =1+\left[p^2-2 q\right]+\left[q^2-2 r p\right]+r^2 \\ & =1+p^2-2 q+q^2-2 r p+r^2 \\ & =(q-1)^2+(r-p)^2 \end{aligned} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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