If $$x^4+\frac{1}{x^4}=62$$, then what is the value of $$x^6+\frac{1}{x^6}$$?

Given : $$x^4+\frac{1}{x^4}=62$$

Adding 2 on both sides,

=> $$x^4+\frac{1}{x^4}+2(x^2)(\frac{1}{x^2})=62+2$$

=> $$(x^2+\frac{1}{x^2})^2=64$$

=> $$(x^2+\frac{1}{x^2})=\sqrt{64}=8$$ ------------(i)

Cubing both sides, we get :

=> $$(x^2+\frac{1}{x^2})^3=(8)^3$$

=> $$x^6+\frac{1}{x^6}+3(x^2)(\frac{1}{x^2})(x^2+\frac{1}{x^2})=512$$

=> $$x^6+\frac{1}{x^6}+3(8)=512$$

=> $$x^6+\frac{1}{x^6}=512-24=488$$

=> Ans - (D)