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.
 A: Hint: Let $a=\sqrt[3]{35-x^3}$. Then you're solving $ax(a+x)=30$.
$(a+x)^3=\left(a^3+x^3\right)+3ax(a+x)=35+3\cdot 30=5^3$.
$\iff a+x=5$. Then $(5-x)x(5)=30$. Solve this quadratic equation.
A: HINT:
$$x^3+35-x^3+3x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3})=35+3\cdot30 $$
$$(x+\sqrt[3]{35-x^3})^3=5^3$$
Assuming  $x$  to be real, $$x+\sqrt[3]{35-x^3}=5\iff\sqrt[3]{35-x^3}=5-x$$
Take cube in both sides
A: 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 $$x=2,2\omega,2\omega^2,3,3\omega,3\omega^2$$

A: I get the answer by factorizing 30 as 2.3.5
now x=3
I don't know if there is any other way .lets check again
A: See you need $6.5$ in reduced form . So we can just manipulate an get solution . So we need a cubic term to get real number so $x\sqrt{35-x^3}=6$ and $x=3$ gives you $6$ also $x+\sqrt{35-x^3}=5$ where $3$ is the solution thus we have one real answer which is 3 other way is solving by Vieta method.
