Roots of a polynomial : removing fractional powers

Question

The quadratic equation $x^2-5x+7=0$ has roots $\alpha$ and $\beta$. Find the quadratic equation with integer coefficents that has roots $\alpha^3$ and $\beta^3$.

I know how to get the answer by considering sums/products of roots but I'm also aware that you can get the answer by use of a substitution $u=x^{\frac{1}{3}}$. This leads to

$u^{\frac{2}{3}}-5u^{\frac{1}{3}}+7=0$

How can you get from here to a quadratic equation?

Answer 1

From

$$u^{\frac{2}{3}}-5u^{\frac{1}{3}}+7=0 \iff (u^{\frac{2}{3}}-5u^{\frac{1}{3}})^3=(-7)^3\iff u^2-125u+343-15u^\frac53+75u^\frac43=0$$

and

$$u^{\frac{2}{3}}=5u^{\frac{1}{3}}-7\implies u^\frac53=5u^\frac43-7u\implies -15u^\frac53=-75u^\frac43+105u$$

then

$$u^2-20u+343=0$$

Answer 2

The quadratic you want is $$(u^{\frac{2}{3}}-5u^{\frac{1}{3}}+7)(u^{\frac{2}{3}}\omega^2-5u^{\frac{1}{3}}\omega+7)(u^{\frac{2}{3}}\omega^4-5u^{\frac{1}{3}}\omega^2+7)\ ,$$ where $$\omega=e^{2\pi i/3}\ .$$ After a good deal of messy algebra this works out to $u^2-20u+343$.

To see a bit more clearly where this comes from, put the original equation back in terms of $x$: it is $$p(x)p(x\omega)p(x\omega^2)\ ,$$ where $p(x)=x^2-5x+7$ is your original polynomial.

Answer 3

Using that $\,u^{\frac{2}{3}}-5u^{\frac{1}{3}}=-7\,$, and the identity $\,(a-b)^3=a^3-b^3-3ab(a-b)\,$:

$$(-7)^3=\left(u^{\frac{2}{3}}-5u^{\frac{1}{3}}\right)^3=u^2-125u-3 \cdot 5u \cdot(-7) \;\;\iff\;\; u^2 - 20u + 343 = 0$$