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

The correct option is D

$x^3+6x^2-36x+27=0$

Explanation for correct option:

Solving to find the equation:

Given equation is $x^3+2x^2-4x+1=0$.

Let the root of this equation be '$\alpha$'

By the given condition, the root of the required equation is '$3\alpha$'

$\therefore {\alpha }^3+2{\alpha }^2-4\alpha +1=0$

So, we can say $x=3\alpha \Rightarrow \alpha =\frac{x}{3}$

Substituting $\alpha$ in the above equation

$\begin{array}{rcl}{\left(\frac{x}{3}\right)}^3+2{\left(\frac{x}{3}\right)}^2-4\left(\frac{x}{3}\right)+1& =& 0\\ \frac{x^3}{27}+\frac{2}{9}{x}^2-\frac{4}{3}x+1& =& 0\\ \frac{x^3+6x^2-36x+27}{27}& =& 0\\ x^3+6x^2-36x+27& =& 0\end{array}$

Hence, option (D) is the correct answer