Q56 / 160 · AP EAMCET 2008 · Mathematics
The cubic equation whose roots are thrice to each of the roots of $x^3+2 x^2-4 x+1=0$ is
- A$x^3-6 x^2+36 x+27=0$
- B$x^3+6 x^2+36 x+27=0$
- C$x^3-6 x^2-36 x+27=0$
- Correct answerD$x^3+6 x^2-36 x+27=0$
Solution
Given equation is
$$
x^3+2 x^2-4 x+1=0
$$
Let $\alpha, \beta$ and $\gamma$ be the roots of the given equation
$$
\begin{aligned}
\therefore & \alpha+\beta+\gamma & =-2, \alpha \beta+\beta \gamma+\gamma \alpha=-4 \\
\text { and } & \alpha \beta \gamma & =-1
\end{aligned}
$$
Let the required cubic equation has the roots $3 \alpha, 3 \beta$ and $3 \gamma$.
$$
\begin{aligned}
\Rightarrow \quad 3 \alpha+3 \beta+3 \gamma & =-6 \\
3 \alpha \cdot 3 \beta+3 \beta \cdot 3 \gamma+3 \gamma \cdot 3 \alpha & =-36 \\
3 \alpha \cdot 3 \beta \cdot 3 \gamma & =-27
\end{aligned}
$$
$\therefore$ Required equation is
$$
\begin{aligned}
x^3-(-6) x^2+(-36) x-(-27) & =0 \\
\Rightarrow \quad x^3+6 x^2-36 x+27 & =0
\end{aligned}
$$
Hence, the correct answer is option: 'D'.