# Finding a quadratic equation using roots

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 $$(x-x_1^3)(x-x_2^3)$$

$$x_{1/2}=\large{-b\pm{\sqrt{4ac}}\over2a}$$

and from here:

\begin{align}x_1^3&=\bigg({-b+{\sqrt{4ac}}\over2a}\bigg)^3\\&={(\sqrt{4ac}-b)^2(\sqrt{4ac}-b)\over8a^3}\\&={(4ac-2b\sqrt{4ac}+b^2)(\sqrt{4ac}-b)\over8a^3}\\&={4ac\sqrt{4ac}-4abc-8abc-2b^2\sqrt{4ac}+b^2\sqrt{4ac}-b^3\over8a^3}\\&={4ac\sqrt{4ac}-12abc-b^2\sqrt{4ac}-b^3\over8a^3}\end{align}

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.

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

HINT

We have

$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$

$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$

and

$$x_1^3x_2^3=(x_1x_2)^3$$

$$x_1^3+x_2^3=?$$

• 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 Nov 8 '18 at 15:20
• @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.
– user
Nov 8 '18 at 15:24
• Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think Nov 8 '18 at 15:27
• @Aleksa Yes I obtain the same result!
– user
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 $$x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$$