Question : If $a^2+\frac{1}{a^2}=\frac{7}{3}$, then what is the value of $\left(a^3-\frac{1}{a^3}\right)?$
Option 1: $\frac{5}{3 \sqrt{3}}$
Option 2: $\frac{10}{3 \sqrt{3}}$
Option 3: $\frac{7}{3 \sqrt{3}}$
Option 4: $\frac{8}{3 \sqrt{3}}$
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Correct Answer: $\frac{10}{3 \sqrt{3}}$
Solution :
Given: $a^2+\frac{1}{a^2}=\frac{7}{3}$
⇒ $(a-\frac{1}{a})^2 +2=\frac{7}{3}$
⇒ $(a-\frac{1}{a})^2=\frac{7}{3} -2$
⇒ $(a-\frac{1}{a})^2=\frac{1}{3}$
⇒ $(a-\frac{1}{a})=\frac{1}{\sqrt3}$
Now, $(a-\frac{1}{a})^3=a^3-\frac{1}{a^3}-3×a×\frac{1}{a}(a-\frac{1}{a})$
⇒ $(\frac{1}{\sqrt 3})^3=a^3-\frac{1}{a^3}-3(\frac{1}{\sqrt 3})$
⇒ $\frac{1}{3\sqrt 3}=a^3-\frac{1}{a^3}-\sqrt3$
⇒ $\frac{1}{3\sqrt 3}+\sqrt3=a^3-\frac{1}{a^3}$
⇒ $\frac{1+9}{3\sqrt 3}=a^3-\frac{1}{a^3}$
⇒ $a^3-\frac{1}{a^3} = \frac{10}{3\sqrt 3}$
Hence, the correct answer is $\frac{10}{3\sqrt 3}$.
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