Question

If α, β, γ are the roots of $x^3$ + 3x + 2 = 0. Find the equation whose roots are ${\alpha }^3$, ${\beta }^3$, ${\gamma }^3$. Also find the value of ${\alpha }^3$ + ${\beta }^3$ + ${\gamma }^3$ - 3αβγ.

• A. $x^3$ - 9x + 6 $x^2$ + 3 = 0, 0
• B. $x^3$ - 39x + 6 $x^2$ + 8 = 0, 0
• C. $x^3$ + 29x + 6 $x^2$ + 8 = 0, 2
• D. $x^3$ - 39x + 6 $x^2$ + 27 = 0, 0

Answer 1

The correct option is B

$x^3$ - 39x + 6 $x^2$ + 8 = 0, 0

Solution: $x^3$ + 3x + 2 = 0

α + β + γ = 0

αβ + βγ + γα = 3

αβγ = -2

If roots are ${\alpha }^3$, ${\beta }^3$ and ${\gamma }^3$

Replace x → $x^{\frac{1}{3}}$

x + 3 $x^{\frac{1}{3}}$ + 2 = 0

x + 2 = -3 $x^{\frac{1}{3}}$

${(x+2)}^3$ = -27x

$x^3$ + 8 + 6$x^2$ + 12x = 27x, $x^3$ - 39x + 6$x^2$ + 8 = 0

Since α + β + γ = 0

${\alpha }^3$ + ${\beta }^3$ + ${\gamma }^3$ = 3αβγ

${\alpha }^3$ + ${\beta }^3$ + ${\gamma }^3$ - 3αβγ = 0