Question

If $\alpha ,\beta and\gamma$ are the roots of the equation $x^3+2x^2+3x+1=0$. Find the equation whose roots are $\frac{1}{{\alpha }^3},\frac{1}{{\beta }^3},\frac{1}{{\gamma }^3}$

• A. $x^3+15x^2-8x+1=0$
• B. $x^3-x^2+8x+2=0$
• C. $x^3+12x^2-7x+1=0$
• D. $x^3-x^2+12x+1=0$

Answer 1

The correct option is C

$x^3+12x^2-7x+1=0$

$x^3+2x^2+3x+1=0$

$x^3+1=-(2x^2+3x)...\left(1\right)$

Cubing on both sides

${(x^3+1)}^3=-{(2x^2+3x)}^3$

$x^9+3x^6+3x^3+1=-[8x^6+27x^3+18x^3(2x^2+3x)]$

$x^3$ + 1 = -(2$x^2$ + 3x) from equation (1).

$\Rightarrow x^9+3x^6+3x^3+1=-[8x^6+27x^3+18x^3(-x^3-1\left)\right]$

Putting $x^3$ = y in this equation

$y^3$ + 3$y^2$ +3y + 1 = $-[8y^2+27y+18y(-y-1)]$

$\Rightarrow y^3-7y^2+12y+1=0$
Its roots will be ${\alpha }^3,{\beta }^3,{\gamma }^3...\left(2\right)$

To get the equation whose roots are $\frac{1}{{\alpha }^3},\frac{1}{{\beta }^3},\frac{1}{{\gamma }^3}$, Change $y$ to $\frac{1}{y}$ in equation (2)

$\Rightarrow \frac{1}{y^3}-\frac{7}{y^2}+\frac{12}{y}+1=0$

Hence required equation in $y$ is
$y^3+12y^2-7y+1=0$
Change the variable to $x$, we have
$\Rightarrow x^3+12x^2-7x+1=0$