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If $$(a+ \frac{1}{a})^{2}=3$$ then the value of $$a^{3}+\frac{1}{a^{3}}$$ is
0
9
$$3\sqrt{3}$$
$$\sqrt{3}$$
$$ a ^ {3} + b ^ {3} = (a+b) ( a^{2} + b^{2} - ab) $$
Substituting b = 1/a, we get the answer as 0
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