It was pointed out in a mathologer video on the cubic formula that $\sqrt[3]{20 + \sqrt{392}} + \sqrt[3]{20 - \sqrt{392}}$ is actually equal to $4$. Is there a series of transformations that can be done to reduce the above expression with radicals to $4$ ?
I realise that writing $x= \sqrt[3]{20 + \sqrt{392}} + \sqrt[3]{20 - \sqrt{392}}$ and cubing both sides gets us to $x^3 = 40 + 6x$ and it's easy to verify that 4 is a solution. Is there a way to arrive at $4 $instead ?
On
OP's solution is the simpler and more direct one, with the following observation.
cubing both sides gets us to $x^3 = 40 + 6x$ and it's easy to verify that $4$ is a solution
It is not enough that $4$ is a solution to prove that $x=4$. What matters here is that $4$ is the only real solution, so the real root $x$ must be equal to $4$.
Mostly for fun, here is another way, to derive the same result. Let $\,a,b = 20 \pm \sqrt{392}\,$, then $\,a + b = 40\,$ and $\,ab = 8\,$, so $\,a,b\,$ are the roots of the quadratic $\,t^2 - 40 t + 8\,$. It follows that $\,\sqrt[3]{a}, \sqrt[3]{b}\,$ are among the roots of the sextic resulting from the substitution $\,t = u^3\,$, which very conveniently happens to factor as:
$$ u^6 - 40 u^3 + 8 = (u^2 - 4 u + 2) (u^4 + 4 u^3 + 14 u^2 + 8 u + 4) $$
The quartic factor has no real roots, because:
$$ u^4 + 4 u^3 + 14 u^2 + 8 u + 4 \;=\; (u+1)^4 + \frac{1}{2} (4 u + 1)^2 + \frac{5}{2} \;\gt\; 0 $$
Therefore the real numbers $\,\sqrt[3]{a}, \sqrt[3]{b}\,$ must be the roots of the quadratic factor $\,u^2 - 4 u + 2\,$, so by Vieta's relations $\,\sqrt[3]{a} + \sqrt[3]{b} = 4\,$.
Your way is very fine, as an alternative we have that by
and then
$$(u+\sqrt v)(u-\sqrt v)=u^2-v=\sqrt[3]{(20 + \sqrt{392})(20 - \sqrt{392})}=\sqrt[3]8=2$$
that is $u^2-v=2$, therefore
$$20 + \sqrt{392}=\left(u+\sqrt {u^2-2}\right)^3=4u^3-6u+(4u^2-2)\sqrt {u^2-2}$$
that is
then $v=2$ and
which finally leads to