If $$(a+\frac{1}{a})^2=3$$, then the value of $$a^6-\frac{1}{a^6}$$ will be

Expression : $$(a+\frac{1}{a})^2=3$$

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

=> $$a^2+\frac{1}{a^2}=3-2=1$$

Squaring both sides, we get :

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

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

=> $$a^4+\frac{1}{a^4}=1-2=-1$$ -----------(i)

$$\therefore$$ $$a^6-\frac{1}{a^6}=(a^2)^3-(\frac{1}{a^2})^3$$

Using, $$x^3-y^3=(x-y)(x^2+y^2+xy)$$

= $$(a^2-\frac{1}{a^2})\times[a^4+\frac{1}{a^4}+(a^4)(\frac{1}{a^4})]$$

Substituting value from equation (i),

= $$(a^2-\frac{1}{a^2})\times(-1+1)=0$$

=> Ans - (C)