How to go from $u^{2/3}-5u^{1/3}+7=0$ to $u^2-20u+343=0$ I have absolutely no clue how to do this - any hints would be appreciated.
So far, I have tried multiplying each term by a power of u, but this leads nowhere...

Cube both sides of $u^{2/3}=5u^{1/3}-7$ to get
$$
\begin{align}
u^2
&=125u\overbrace{-525u^{2/3}+735u^{1/3}}^{-105u}-343\\
&=20u-343
\end{align}
$$
where $-105u=-105\overbrace{\left(5u^{1/3}-7\right)}^{u^{2/3}}u^{1/3}$
Therefore,
$$
u^2-20u+343=0
$$

In General
Suppose we had $u^{2/3}=au^{1/3}+b$, then
$$
\begin{align}
u
&=au^{2/3}+bu^{1/3}\\
&=a\left(au^{1/3}+b\right)+bu^{1/3}\\
&=\color{#C00}{\left(a^2+b\right)u^{1/3}+ab}\\
u^2
&=\left(a^2+b\right)^2u^{2/3}+2ab\left(a^2+b\right)u^{1/3}+(ab)^2\\
&=\left(a^2+b\right)^2\left(au^{1/3}+b\right)+2ab\left(a^2+b\right)u^{1/3}+(ab)^2\\
&=\color{#C00}{\left(a^2+b\right)\left(a^3+3ab\right)u^{1/3}+b\left(a^4+3a^2b+b^2\right)}
\end{align}
$$
Then, we can combine the two red formulas to get
$$
u^2=\left(a^3+3ab\right)u+b^3
$$

Complex Approach
Let $v=u^{1/3}$ and $\omega=e^{2\pi i/3}$. Then note that
$$
\left(v^2-av-b\right)\left(v^2\omega^2-av\omega-b\right)\left(v^2\omega-av\omega^2-b\right)=v^6-\left(a^3+3ab\right)v^3-b^3
$$
