2024/002) If x = $2^{\frac{1}{3}} + 2^{-\frac{1}{3}}$ , How do you prove that $2x^3-6x = 5$

 Let $a =  2^{\frac{1}{3}}\cdots(1)$

So $a^3 = 2\cdots(2)$ 

And $\frac{1}{a} = 2^{-\frac{1}{3}}\cdots(3)$

And  

$x = a+ \frac{1}{a}\cdots(4)$

Cubing both sides $x^3 = a^3 + \frac{1}{a^3} + 3 * a * \frac{1}{a} ( a + \frac{1}{a})$

Or $x^3 = a^3 + \frac{1}{a^3} + 3 * ( a + \frac{1}{a})$

Or  $x^3 = 2 + \frac{1}{2} + 3 *x$ from (2), 3) and (4)

Or $x^3 =  \frac{5}{2} + 3 *x$

Or $2x^3 = 5 + 6x$

Or $2x^3-6x = 5$

Proved