Factor and Solve x^8 - 80x^4 - 81

Factor x^8 - 80x^4 - 81 over COMPLEX NUMBERS and then solve the equation x^8-80x^4-81=0.

I'm lost in this. I didn't expect complex number questions in my homework since it wasn't covered. please help

——————————————————————————————————————

0 = x⁸ - 80x⁴ - 81            ← Notice that this is a quadratic expression in x⁴
0 = (x⁴)² - 80x⁴ - 81            ← This is like u²-80u-81
                                            Need two factors of -81 that add to -80
                                            Those would be -81 & +1
0 = (x⁴-81)(x⁴+1)            ← Like (u-81)(u+1)
                                        Now, break (x⁴-81) & (x⁴+1) up into
                                        sum and difference of squares
0 = [(x²)²-9²] * [(x²)²+1²]
0 = [(x²-9)(x²+9)] * [(x²)² - i²]
0 = [(x²-9)(x²+9)] * [(x² - i)(x² + i)]

0 = [(x²-9)] * [(x²+9)] * [(x²  -  i)(x²  +  i)]
0 = [(x²-3²)] * [(x²+3²)] * [(x²  -  i)(x²  +  i)]
0 = [(x-3)(x+3)] * [x²-(3i)²] * [(x²  -  i)(x²  +  i)]
0 = [(x-3)(x+3)] * [(x-3i)(x+3i)]  *  [(x² - i)(x² +  i)]
           ➊               ➋              ➌      ➍


Solutions
➊ x = ±3
➋ x = ±3i
➌ x = ±√(i) = ±√(-1)
➍    x² + i
   = x²-(i³)            ← note: i³ = -i
   = x²-(i^³₂̸ )²            ← difference of squares
   = (x-i^³₂̸ )(x+i^³∕₂)
   = (x-[(-1)^½]^³∕₂)(x+[(-1)^½]^³∕₂)            ← substituted √(-1) in for i
   = (x-(-1)^¾)(x+(-1)^¾)
   = (x-∜[(-1)³])(x+∜[(-1)³])
   = (x-∜ ̅-̅1̅ ̚ )(x+∜ ̅-̅1̅ ̚ )
           x = ± ∜ ̅-̅1̅ ̚            ← solution


             ANSWER
x = ±3, ±3i, ±√(-1), ± ∜ ̅-̅1̅ ̚

Hope that's useful.
Have a good one!
——————————————————————————————————————

y = x^4, so your question is now

y² - 80y - 81 = 0

(y - 81)(y + 1) = 0

That is true only if y = 81 or -1

Remember y = x^4, so

x^4 = 81, therefore x = 3

x^4 = -1, this isn't possible even with complex numbers because i^4 = 1

Let x^4 be = a

a^2 -80a -81

( a +1) ( a- 81)

(x^4 +1) ( x^4 -81)

( x^4 +1) ( x+ 3) ( x-3) ( x^2 +9) ANSWER

Good explanation by Bill.