Olympiad corner

Plus two sides of the equation we get: $x^8-28y^2x^6+70x^4y^4-28y^6x^2+y^8+i(8yx^7-56y^3x^5+56y^5x^3-8y^7x)=\frac{\sqrt{3}}{2}+i\frac{1}{2}$ $\Leftrightarrow (x+iy)^8=\frac{\sqrt{3}}{2}+i\frac{1}{2}=cos\frac{\pi }{6}+isin\frac{\pi }{6}$ $\Leftrightarrow x+iy=cos\frac{\frac{\pi}{6}+k2\pi }{8}+isin\frac{\frac{\pi }{6}+k2\pi }{8}(k=0,1,2,...,7)$ This system of equation have 8 pairs: $(x,y)=(cos\frac{\frac{\pi}{6}+k2\pi}{8};sin\frac{\frac{\pi}{6}+k2\pi}{8}),(k=0,1,2,..7)$ So the answer is 8 PAIRS.