I can't figure out this algebra problem, please help?

Definitely a tricky step.

expand 1 + (x^4 / 2 - 1 / 2x^4)^2 =

1 + (x^4 / 2 - 1 / 2x^4)(x^4 / 2 - 1 / 2x^4), now FOIL to get:

1 + x^8 / 4 - x^4/4x^4 - x^4/4x^4 + 1/4x^8 =

1 + x^8/4 - 2x^4/4x^4 + 1/4x^8 =

1 + x^8/4 - 1/2 + 1/4x^8 =

x^8/4 + 1/4x^8 + 1/2, here is were its tough to notice the next step, but if you remember (a + b)^2 = a^2 + 2ab + b^2, you might see that x^8/4 and 1/4x^8 are perfect squares and thier square roots(x^4/2, 1/2x^4) if you multiply them you get 1/4 and two times that is 1/2. so you have a a^2 +2ab + b^2 polynomial which factors to (a + b)^2 or

(x^4/4 + 1/2x^4).

1 + ( X^4 / 2 - 1 / 2X^4 )^2 = ( X^4 / 2 + 1 / 2X^4 )^2

Heathere...

Left side

[1 + [( X^4 / 2) - (1 / 2X^4 )]^2

1 + [X^4(X^4)] / 2X^4 - 1 / 2X^4]^2

1 + [X^4(X^4)] - 1]² / 4X^8

4X^8 + [X^4(X^4)] - 1]² / 4X^8

{4X^8 + (X^8 - 1)(X^8 - 1)}/4X^8

{4X^8 + (X^16 - 2X^8 + 1)} / 4X^8

(X^16 + 2X^8 + 1) / 4X^8 ← left side

Right side

( X^4/ 2 ) + (1/ 2X^4 )^2

[( X^4)(X^4 )/2X^4 + (1/ 2X^4 )]^2

[(X^8 + 1)/2X^4]²

(X^8 + 1)²/(2X^4)²

(X^8 + 1)(X^8 + 1)/4X^8

(X^16 + 2X^8 + 1)/4X^8 ← Right side

Left Side = Right Side ◄ Answer

hope this helps