Q91 / 160 · TS EAMCET 2014 · Mathematics

The cubic equation whose roots are the squares of the roots of $x^3-2 x^2+10 x-8=0$, is

  • Correct answer
    A
    $x^3+16 x^2+68 x-64=0$
  • B
    $x^3+8 x^2+68 x-64=0$
  • C
    $x^3+16 x^2-68 x-64=0$
  • D
    $x^3-16 x^2+68 x-64=0$

Solution

Let, $\alpha, \beta$ and $\gamma$ are the roots of $$ \begin{aligned} & x^3-2 x^2+10 x-8 & =0 \\ \therefore & \alpha+\beta+\gamma & =2, \end{aligned} $$ $$ \alpha \beta+\beta \gamma+\gamma \alpha=10 $$ and $\alpha \beta \gamma=8$ Now, $\alpha^2+\beta^2+\gamma^2$ $$ \begin{aligned} & =(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha) \\ & =(2)^2-2(10)=-16 \end{aligned} $$ $$ \begin{aligned} \alpha^2 \beta^2+ & \beta^2 \gamma^2+\gamma^2 \alpha^2 \\ & =(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2(\alpha+\beta+\gamma)(\alpha \beta \gamma) \\ & =(10)^2-2(2)(8) \\ & =100-32=68 \end{aligned} $$ and $$ \alpha^2 \beta^2 \gamma^2=(8)^2=64 $$ $\therefore$ Required cubic equation is $$ \begin{aligned} & x^3-\left(\alpha^2+\beta^2+\gamma^2\right) x^2+\left(\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2\right) x \\ & -\alpha^2 \beta^2 \gamma^2=0 \\ & \therefore x^3-(-16) x^2+(68) x-64=0 \\ & \Rightarrow \quad x^2+16 x^2+68 x-64=0 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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