1. factoring sixth degrees

Can someone help me out here?

x^6 - 7x^3 - 8 = 0

THere must be a simpler way to factoring this without having to use factor theorem! Someone please help?

2. $x^6-7x^3-8=0\implies(x^3+1)(x^3-8)=0$

Continue.

3. Originally Posted by checkmarks
Can someone help me out here?

x^6 - 7x^3 - 8 = 0

THere must be a simpler way to factoring this without having to use factor theorem! Someone please help?
It's a quadratic in $y=x^3$

RonL

4. ahh i thought of that.
but i just assumed an exponent raised to an exponent, you would multiply...so (x^3)(x^3) = x^9
shows you how terrible i am at simple math

thanks guys