Question 60

If $$(a+\frac{1}{a})^2=3$$, the value of $$a^3+\frac{1}{a^3}$$ is

Given : $$(a+\frac{1}{a})^2=3$$

=> $$(a+\frac{1}{a})=\sqrt3$$ -----------(i)

Cubing both sides, we get :

=> $$(a+\frac{1}{a})^3=(\sqrt3)^3$$

=> $$a^3+\frac{1}{a^3}+3(a)(\frac{1}{a})(a+\frac{1}{a})=3\sqrt3$$

=> $$a^3+\frac{1}{a^3}+3(\sqrt3)=3\sqrt3$$     [Using (i)]

=> $$a^3+\frac{1}{a^3}=3\sqrt3-3\sqrt3=0$$

=> Ans - (A)

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