If $x_1$ and $x_2$ are the roots of $$ax^2+bx+c=0$$ then $x_1^3$ and $x_2^3$ are the roots of which equation?

So I tried by solving this for $x_{1/2}$ so I could change it in $(x-x_1^3)(x-x_2^3)$


and from here:


but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.

  • $\begingroup$ Hint: Just replace $x$ with $\sqrt[3] x$ in the given quadratic to obtain required quadratic. $\endgroup$ – Darkrai Nov 8 '18 at 15:13
  • $\begingroup$ @Manthanein: that would not be a quadratic. $\endgroup$ – Martin Argerami Nov 8 '18 at 23:36


We have






  • $\begingroup$ Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think $\endgroup$ – Aleksa Nov 8 '18 at 15:20
  • 1
    $\begingroup$ @Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms. $\endgroup$ – gimusi Nov 8 '18 at 15:24
  • $\begingroup$ Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think $\endgroup$ – Aleksa Nov 8 '18 at 15:27
  • 1
    $\begingroup$ @Aleksa Yes I obtain the same result! $\endgroup$ – gimusi Nov 8 '18 at 15:32

Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.

The roots of the polynomial $$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$ are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.