Fast way to solve $4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}$

Question

The question is this:

$4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}$

For some reason, I keep on getting 289/3, even though it is the wrong answer. This is from a timed test, and my way is wrong and extremely slow.

Answer 1

If $\sqrt[3]{x-10}=t$, then the equation is

$$4+t=\sqrt[3]{t^3+20}.$$

So $$64+48t+12t^2+t^3=t^3+20,$$ $$\iff 12t^2+48t+44=0,$$ which gives $ t=(-24\pm \sqrt{48})/12=-2\pm \sqrt{3}/3$ and $$x=t^3+10=\pm 7.12065332001...$$

Answer 2

Taking the cube on both sides,
$\implies 64 = 10 - 3(4)(x^2 - 100)^{1/3}$.

Note that $(x+10)^{1/3} - (x-10)^{1/3} = 4$

Thus, we get $x^2 = - \frac{1369}{27}$.

Hence, the roots are :
$-\frac{37\sqrt3}{3}$ & $\frac{37\sqrt3}{3}$.

Answer 3

Use this formula $$\boxed{(a-b)^3 = a^3-3ab(a-b)-b^3}$$

$$4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}\;\;\;/^3$$

$$64 = x+10 -3\sqrt [3]{(x+10)(x-10)}(\underbrace{\sqrt[3] {x+10}-\sqrt[3] {x-10}}_{4})-x+10$$

So $$11 = -3\sqrt [3]{x^2-100}\implies x = \pm\sqrt{{-11^3\over 27}+100}$$