5 12) \( x^{6}-4 x^{4}-16 x^{2}+64=0 \)

To solve the equation \( x^{6}-4 x^{4}-16 x^{2}+64=0 \), we can start by making a substitution. Let \( y = x^2 \). This transforms our equation into a cubic one: \( y^{3} - 4y^{2} - 16y + 64 = 0 \). Next, we can attempt to find rational roots using the Rational Root Theorem or try factoring. Looking at possible integer factors of 64, we can test \( y = 4 \) and find that it works: \( 4^3 - 4(4^2) - 16(4) + 64 = 64 - 64 - 64 + 64 = 0 \). Thus, \( y - 4 \) is a factor. To find the other factors, we can perform synthetic division or polynomial long division on \( y^{3} - 4y^{2} - 16y + 64 \) by \( y - 4 \). After division, we find that \( y^{3} - 4y^{2} - 16y + 64 = (y - 4)(y^2 + 0y - 16) \). Setting \( y^2 - 16 = 0 \), we can solve for \( y \) and find two more roots: \( y = 4 \) and \( y = -4 \). Recalling \( y = x^2 \), we can substitute back to find \( x \). From \( y = 4 \), we have \( x^2 = 4 \) which gives \( x = 2 \) or \( x = -2 \). From \( y = -4 \), since no real numbers squared yield a negative, we discard these solutions. Thus, the real solutions for \( x^{6}-4 x^{4}-16 x^{2}+64=0 \) are \( x = 2, -2 \).