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