If x^2-√11 x+1=0, then ≤ft(x^3+x^-3)=

If $x^2-\sqrt{11} x+1=0$

divide by x on both sides,

x + \(\frac{1}{x}\) = \(\sqrt {11}\)

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

 $x^3 +\frac{1}{x^3}$ = (\(\sqrt {11}\))³ - 3 × \(\sqrt {11}\) = $8 \sqrt{11}$