Solve for reals $[2022+x(2+\sqrt{x})][111-x(2+\sqrt{x})]=-52^3$

Question

Solve in real numbers:

$$\sqrt[3]{2022+x(2+\sqrt{x})}+\sqrt[3]{111-x(2+\sqrt{x})}=9$$

Cubing both sides, we get$$ 2022+111+3\sqrt[3]{2022+x(2+\sqrt{x})}\sqrt[3]{111-x(2+\sqrt{x})}\Big(\sqrt[3]{2022+x(2+\sqrt{x})}+\sqrt[3]{111-x(2+\sqrt{x})}\Big)=9^3\implies $$ $$ 1404+3\times 9\sqrt[3]{2022+x(2+\sqrt{x})}\sqrt[3]{111-x(2+\sqrt{x})}=0$$ $$\implies \sqrt[3]{2022+x(2+\sqrt{x})}\sqrt[3]{111-x(2+\sqrt{x})}=-52\implies [2022+x(2+\sqrt{x})][111-x(2+\sqrt{x})]=-52^3$$

$$ \implies -4 x^{5/2}- 1911 x^{3/2} - x^3 - 4 x^2 - 3822 x + 224442 = -140608\implies 4 x^{5/2}+ 1911 x^{3/2} + x^3 +4 x^2 + 3822 x =365050$$

Take $\sqrt{x}=y\implies $ $$4y^5+1911 y^3 + y^6 +4y^4 + 3822y^2 =365050\implies y^2(y^4+4y^3+4y^2+1911y+3822)=365050\implies y^2(y + 2) (y^3 + 2 y^2 + 1911)=365050$$

Solutions or help?

Answer 1

I would use a substitution to keep degrees low, like so: with $a^3=2022+x(2+\sqrt x), b^3 = 111-x(2+\sqrt x)$, we get the simpler system $a^3+b^3=2133, a+b=9$ to solve first.
$\implies a^2-ab+b^2 = 2133/9 = 237 \implies 237 = (a+b)^2-3ab \implies ab = -52$

Thus $a, b$ are roots of $t^2-9t-52 = (t-13)(t+4)=0$. Now, as $x$ needs to be non-negative, we have $a> b$, so $a=13, b=-4$.

Now $a^3 = 2022+x(2+\sqrt x) \implies x(2+\sqrt x)=175=25\cdot (2+5)\implies x=25$, as $x(2+\sqrt x)$ is obviously an increasing function.

Answer 2

This is more of a comment than an answer, but I wanted to show a method.

If $a,b,c$ are the roots of the cubic $z^3+pz-q=0$ (so that $a+b+c=0$ and $q=abc$) then $$0=(a^3+pa-q)+(b^3+pb-q)+(c^3+pc-q)=a^3+b^3+c^3-3q$$

so that $a^3+b^3+c^3=3abc$

The operative condition here is that $a+b+c=0$.

This has an application where where $a+b=-c$ and $a$ and $b$ are cube roots with some element of conjugacy beneath the cube root, as the conjugate elements sometimes multiply nicely together.

You may be able to apply this here.

This is actually related to the solution of the cubic in an interesting way (the the formula for the roots of the general 'depressed' cubic is expressed as the sum of cube roots of conjugate expressions - and $a^3+b^3+c^3=3abc$ is a cubic in $c$ which is satisfied by $c$. Any cubic can be reduced quite easily to 'depressed' form)