If $(x+\frac{1}{x})^3=27$, then what is the value of $(x^2 + \frac{1}{x^2})$ ? Given that x is real.

7

We know that,

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

If $(x+\frac{1}{x})^3=27$,

then what is the value of $(x^2 + \frac{1}{x^2})$ = ?

If $(x+\frac{1}{x})^3=27$

then, If $(x+\frac{1}{x})=\sqrt[3]{27}$ = 3

So, $(x^2 + \frac{1}{x^2})$ = 3² – 2 = 7