Question : When $2x+\frac{2}{x}=3$, then the value of ($x^3+\frac{1}{x^3}+2)$ is:
Option 1: $\frac{2}{7}$
Option 2: $\frac{7}{8}$
Option 3: $\frac{7}{2}$
Option 4: $\frac{8}{7}$
Correct Answer: $\frac{7}{8}$
Solution :
Given: $2x+\frac{2}{x}=3$
⇒ $x+\frac{1}{x}=\frac{3}{2}$ (Dividing both sides by 2)
Now, $(x+\frac{1}{x})^3=(\frac{3}{2})^3$ (Taking cube on both sides)
⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=\frac{27}{8}$
⇒ $x^3+\frac{1}{x^3}+3(\frac{3}{2})=\frac{27}{8}$
⇒ $x^3+\frac{1}{x^3}=\frac{27}{8}–\frac{9}{2}$
⇒ $x^3+\frac{1}{x^3}+2=\frac{27}{8}–\frac{9}{2}+2=\frac{27–36+16}{8}=\frac{7}{8}$
Hence, the correct answer is = $\frac{7}{8}$.
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