Solve $ \left(\sqrt[3]{4-\sqrt{15}}\right)^x+\left(\sqrt[3]{4+\sqrt{15}}\right)^x=8 $ [closed]
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,
$$A^{x/3}=\frac{8\pm\sqrt{64-4}}{2}=\frac{8\pm2\sqrt{15}}{2}=4\pm\sqrt{15}\\ \implies (4+\sqrt{15})^{x/3}=(4+\sqrt{15})^{\pm 1}\\ \implies \frac{x}{3}=\pm 1 \implies x=\pm 3$$
Hint: $$\sqrt[3]{4-\sqrt{15}}\cdot \sqrt[3]{4+\sqrt{15}}=1$$ from here we get $$t^x+\frac{1}{t^x}=8$$