Question : If $\sqrt{x}+\frac{1}{\sqrt{x}}=3$, then the value of $x^3+\frac{1}{x^3}$ is:
Option 1: 326
Option 2: 322
Option 3: 324
Option 4: 422
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Correct Answer: 322
Solution :
Given: $\sqrt{x}+\frac{1}{\sqrt{x}}=3$ (equation 1)
Use the formulas,
If $x+\frac{1}{x}=n$. Then, $x^2+\frac{1}{x^2}=n^2–2$ and $x^3+\frac{1}{x^3}= (x+\frac{1}{x})^3 -3(x+\frac{1}{x}) = n^3–3n$.
Squaring the equation 1, we get,
$(\sqrt{x}+\frac{1}{\sqrt{x}})^2=3^2$
⇒ $x+\frac{1}{x}+2=9$
⇒ $x+\frac{1}{x}=7$ (equation 2)
The value of $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})=7^3-3\times 7=343–21=322$
Hence, the correct answer is 322.
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