Question : If $x^2+\frac{1}{x^2}=7$, then what is the value of $x^3+\frac{1}{x^3}$?
Option 1: 9
Option 2: 18
Option 3: 27
Option 4: 36
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Correct Answer: 18
Solution :
$x^2+\frac{1}{x^2}=7$
Adding 2 on both sides, we get,
⇒ $(x+\frac{1}{x})^2 - 2 × x × \frac{1}{x} = 7+2$
⇒ $(x+\frac{1}{x})^2 = 9$
⇒ $x+\frac{1}{x} = 3$
cubing on both sides, we get,
⇒ $(x+\frac{1}{x})^3 = 3^3$
⇒ $x^3 + \frac{1}{x^3} + 3 × x × \frac{1}{x}(x+\frac{1}{x}) = 27$
⇒ $x^3 + \frac{1}{x^3} + 3 × 3 = 27$
$\therefore x^3 + \frac{1}{x^3} =18$
Hence, the correct answer is 18.
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