If (a+frac{1}{a})^{2}=3, then a^{3}+frac{1}{a | vedclass

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(D) Given that $(a+\frac{1}{a})^{2}=3$.Taking the square root on both sides,we get $a+\frac{1}{a}=\sqrt{3}$.To find $a^{3}+\frac{1}{a^{3}}$,we use the

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(D) Given that $(a+\frac{1}{a})^{2}=3$.Taking the square root on both sides,we get $a+\frac{1}{a}=\sqrt{3}$.To find $a^{3}+\frac{1}{a^{3}}$,we use the algebraic identity $(x+y)^{3} = x^{3} + y^{3} + 3xy(x+y)$.Substituting $x=a$ and $y=\frac{1}{a}$,we get $(a+\frac{1}{a})^{3} = a^{3} + \frac{1}{a^{3}} + 3(a)(\frac{1}{a})(a+\frac{1}{a})$.This simplifies to $(a+\frac{1}{a})^{3} = a^{3} + \frac{1}{a^{3}} + 3(a+\frac{1}{a})$.Substituting the value $a+\frac{1}{a}=\sqrt{3}$ into the equation:$(\sqrt{3})^{3} = a^{3} + \frac{1}{a^{3}} + 3(\sqrt{3})$.$3\sqrt{3} = a^{3} + \frac{1}{a^{3}} + 3\sqrt{3}$.Subtracting $3\sqrt{3}$ from both sides,we get $a^{3} + \frac{1}{a^{3}} = 0$.

If $(a+\frac{1}{a})^{2}=3,$ then $a^{3}+\frac{1}{a^{3}}=?$

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