Answer :
the fully factorized expression is:
[tex]\[ {(x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1)} \][/tex]
let's factorize the expression [tex]\( x^6 - 7x^3 - 8 \)[/tex].
First, let's introduce a substitution. Let [tex]\( y = x^3 \)[/tex]. So, the expression becomes [tex]\( y^2 - 7y - 8 \).[/tex]
Now, we need to factorize the quadratic expression [tex]\( y^2 - 7y - 8 \)[/tex]. We can do this by finding two numbers that multiply to give -8 and add to give -7. These numbers are -8 and 1.
Rewrite -7y as -8y + y and then factor by grouping:
[tex]\[ y^2 - 8y + y - 8 \][/tex]
Factor by grouping:
[tex]\[ y(y - 8) + 1(y - 8) \][/tex]
Factor out the common factor (y - 8):
(y - 8)(y + 1)
Now, substitute back [tex]\( x^3 \) for \( y \):[/tex]
[tex]\[ (x^3 - 8)(x^3 + 1) \][/tex]
Notice that [tex]\( x^3 - 8 \)[/tex] is a difference of cubes, which can be factored as [tex]\( (x - 2)(x^2 + 2x + 4) \).[/tex] And [tex]\( x^3 + 1 \)[/tex] is a sum of cubes, which can be factored as [tex]\( (x + 1)(x^2 - x + 1) \).[/tex]
So, the fully factorized expression is:
[tex]\[ {(x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1)} \][/tex]
The complete question is:
Factorize the expression [tex]\( x^6 - 7x^3 - 8 \)[/tex] ?
