<s>[INST] Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
 [/INST] Setting $t=\sqrt[3]{35-x^3}$, we have
$$tx(t+x) = 30$$
and
$$t^3+x^3=35 \implies (t+x)(t^2+x^2-tx)=35$$
This gives us
$$\dfrac{t^2+x^2-tx}{tx} = \dfrac76 \implies 6t^2-13tx+6x^2 = 0 \implies 6t^2 - 9tx - 4tx + 6x^2 = 0$$
This gives us
$$3t(2t-3x)-2x(2t-3x) = 0 \implies (2t-3x)(3t-2x) = 0 \implies t = \dfrac{2x}3, \dfrac{3x}2$$
Trust you can finish it from here easily.

We then obtain $$t^3 = \dfrac8{27}x^3, \dfrac{27}8x^3 \implies 35-x^3 = \dfrac8{27}x^3, \dfrac{27}8x^3 \implies x^3=8,27$$This gives us $$x=2,2\omega,2\omega^2,3,3\omega,3\omega^2$$

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