Solve $ \left(\sqrt[3]{4-\sqrt{15}}\right)^x+\left(\sqrt[3]{4+\sqrt{15}}\right)^x=8 $

Question

I don't know what can I substitute for $x$ so that equation becomes satisfied. Any assistance will be greatly valued.

Thanks!

Answer 1

Use $A=4+\sqrt{15}$. That'll give you $A^{-1}=\dfrac{1}{4+\sqrt{15}}=\dfrac{4-\sqrt{15}}{16-15}=4-\sqrt{15}$

The equation then becomes,

$$A^{x/3}+A^{-x/3}=8\implies (A^{x/3})^2-8A^{x/3}+1=0$$

Solve this quadratic in terms of $A^{x/3}$ using the quadratic formula and you'll get,

$$A^{x/3}=\frac{8\pm\sqrt{64-4}}{2}=\frac{8\pm2\sqrt{15}}{2}=4\pm\sqrt{15}\\ \implies (4+\sqrt{15})^{x/3}=(4+\sqrt{15})^{\pm 1}\\ \implies \frac{x}{3}=\pm 1 \implies x=\pm 3$$

Answer 2

Hint: $$\sqrt[3]{4-\sqrt{15}}\cdot \sqrt[3]{4+\sqrt{15}}=1$$ from here we get $$t^x+\frac{1}{t^x}=8$$

Answer 3

$$4-\sqrt{15}=t^3,\sqrt{15}=4-t^3$$ $$4+\sqrt{15}=8-t^3$$ $$(\sqrt[3]{8-t^3})^x=8-t^x$$ $$(8-t^3)^{\frac{x}{3}}=8-t^x$$ $$(8-t^3)^x=(8-t^x)^3$$ $x=3$ obviousley is a solution