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.
• B.
• C.
• D.

Answer 1

The correct option is C

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

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

Cubing on both sides

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

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

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

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

Putting $x^3$ = y in this equation

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

$y^3$ - 7$y^2$ +12y + 1 = 0 {its root ${\alpha }^3$, ${\beta }^3$, ${\gamma }^3$}_____________(2)

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

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