Question

The cubic equation whose roots are thrice to each of the roots of $x^3 + 2x^2 - 4x + 1 = 0$ is

• A. $x^3 - 6x^2 + 36x + 27 = 0$
• B. $x^3 + 6x^2 + 36x + 27 = 0$
• C. $x^3 - 6x^2 - 36x + 27 = 0$
• D. $x^3 + 6x^2 - 36x + 27 = 0 $

Answer 1

Answer: (D)

Given equation is
$x^{3}+2 x^{2}-4 x+1=0$
Let $\alpha, \beta$ and $\gamma$ be the roots of the given equation
$\therefore \alpha+\beta+\gamma=-2, \alpha \beta+\beta \gamma+\gamma \alpha=-4$ and $\alpha \beta \gamma=-1$
Let the required cubic equation has the roots $3 \alpha, 3 \beta$ and $3 \gamma$.
$\Rightarrow 3 \alpha+3 \beta+3 \gamma=-6$,
$3 \alpha .3 \beta+3 \beta+3 \gamma+3 \gamma .3 \alpha=36$
and $3 \alpha .3 \beta .3 \gamma=-27$
$\therefore $ Required equation is
$x^{3}-(-6) x^{2}+(-36) x-(-27)=0$
$\Rightarrow x^{3}+6 x^{2}-36 x+27=0$