Just give me one more chance to make it right!

Opening the bracket and rearranging terms we get :

${x}^{2}=x+1$ (i) Multiplying both sides by x we get

${x}^{3}={x}^{2}+x$ Using (i) we get

${x}^{3}=(x+1)+x=2x+1$ Squaring both sides to get

${x}^{6}=4{x}^{2}+4{x}+1$ Again using (i) we get

$\boxed{{x}^{6}=4(x+1)+4x+1=8x+5}$

Of course one can generalise the fact(and prove it by induction) that :

${x}^{n}={F}_{n+1}x+{F}_{n}$ where ${F}_{n}$ is the $nth$ term of fibbonaci series