how can i factor this completely?

Question

x^6 - 7x^3 - 8

thanks, i have trouble factoring trinomials that are above the third power. a solution might help, thank you

Answer 1

x^6-7x^3-8
=x^3-1-7x^3-7
=(x^6-1)-7(x^3+1)
={(x^3)^2-(1)^2}-7(x^3+1)
=(x^3+1)(x^3-1)-7(x^3+1)
={(x)^3+(1)^3}(x^3-1)-7(x^3+1)
=(x^3+1)(x^3-1-7)
=(x+1)(x^-x+1)(x^3-8)
=(x+1)(x^2-x+1){(x)^3-(2)^3}
=(x+1)(x^2-x+1)(x-2)(x^2+2x+4)
A special note to Adefolahan
You must be the first boy of your class and very proficient in Mathematics,but I think you have not so far gone through Factor Theorem,which states that if in a polynomial expression containing x,x is substituted by h and the value of the expression becomes zero,then x-h is a fator of the expression.
In the expression x^6-7x^3-8,if x is substituted by -1,the value of the expression becomes 1+7-8=0.Therefore ,be sure of it that x+1 is a factor of the expression
Similarly,if x is replaced by 2,the value of the expression becomes 64-56-8=0 and hence x-2 is also a factor of the expression.
Please ask your teacher to explain the theorem to you and till such time refrain from making such abrupt remarks

Answer 2

This a quadratic in x³. So, let u = x³.
Then u²-7u-8 = (u-8)(u+1)
Converting back to x³, we get
x^6 - 7x^3 - 8 = (x³-8)(x³+1)
Now the first term is the difference of 2 cubes
and the last is the sum of 2 cubes, so our final result is
x^6 - 7x^3 - 8 = (x-2)(x²+2x+4)(x+1)(x²-x+1).

Answer 3

x^6 = (x^3)^2
-7x^3 = -7(x^2)x


(x^3)^2 - 7(x^2)x - 8

two numbers that multiply to -8 and add to -7

the two numbers are: -8 and 1
(x^3 - 8) (x^3 + 1)

the expression are difference of perfect cubes and the sum of perfect cubes, use this formula:
a^3 - b^3 = (a - b) (a^2 + ab + b^2)
a^3 + b^3 = (a + b) (a^2 - ab + b^2)

(x^3 - 2^3) (x^3 +1^3)

use the formula above and you'll get

(x - 2) (x^2 + 2x + 4) (x + 1) (x^2 - x + 1)