Question

If $\alpha ,and\beta$ are the roots of $x^2+px+q=0$ , the quadratic equation having roots ${\alpha }^3,{\beta }^3$ is given by

• A. $x^2+(3pq-p^3)x-q^3=0$
• B. $p{x}^2-(3pq+p^3)x+q^3=0$
• C. $p{x}^2+(3pq-p^3)x-q^3=0$
• D. $x^2+(p^3-3pq)x+q^3=0$

Answer 1

The correct option is D $x^2+(p^3-3pq)x+q^3=0$

Solution : $\alpha ,\beta$ are the roots of $x^2+px+q=0$
$\alpha +\beta =-p$
$\alpha \beta =q$
$(x-{\alpha }^3)(x-{\beta }^3)=0$
$x^2-({\alpha }^3+{\beta }^3)x+{\alpha }^3{\beta }^3=0$
value of ${\alpha }^3+{\beta }^3=(\alpha +\beta )({\alpha }^2+{\beta }^2-\alpha \beta )$
$=(-p)(p^2-3q)$
$x^3+p(p^2-3q)x+q^3=0$
$x^3-(3pq-p^3)x+q^3=0$