We have in this case the difference of Perfect Cubes since we have:
[tex]x^3-8=x^3-2^3[/tex]We know that this case can be factored as follows:
[tex]a^3-b^3=(a-b)\cdot(a^2+ab+b^2)[/tex]If we have that:
a = x
b = 2
Then, we have:
[tex](x-2)\cdot(x^2_{}+2x+2^2)=(x-2)\cdot(x^2+2x+4)[/tex]Therefore, the factored form of the perfect cube (for difference) is:
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