\(\left(cos^4x-sin^4x\right)^2=\left(cos^2x-sin^2x\right)^2\left(cos^2x+sin^2x\right)^2=cos^22x\)
\(\Rightarrow cos^22x=\frac{1}{3}\Rightarrow\frac{1+cos4x}{2}=\frac{1}{3}\Rightarrow cos4x=-\frac{1}{3}\)
\(cos8x=2cos^24x-1=2.\left(-\frac{1}{3}\right)^2-1=-\frac{7}{9}\)