Question

If $x^2-4x+1=0$,, then what is the value of $x^3+\frac{1}{x^3}$?

• A. $44$
• B. $48$
• C. $52$
• D. $64$

Answer 1

The correct option is C $52$

$(x^3+\frac{1}{x^3})=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)$

$\Rightarrow (x+\frac{1}{x})[{(x+\frac{1}{x})}^2-3]$

As $x^2-4x+1=0$

$\Rightarrow x+\frac{1}{x}-4=0$

$\Rightarrow x^3+\frac{1}{x^3}=4[4^2-3]=52$