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. $\endgroup$ – Digamma 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
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    $\begingroup$ @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
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    $\begingroup$ @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 $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$


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