how do you factorise x^3 -125?

x^3 -125=(x-5)(x^2+5x+25) answer//

Use the difference of cubes formula, which says that

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

So in your case, the "a" = "x" and the "b" = "5" (since 5^3=125)

Therefore,

x^3 - 125
= x^3 - 5^3
= (x-5)(x^2 + 5x + 25)

[As a check, If you expand that, you'll get back to x^3-125.]

f(x) = x^3 -125
f(5) = 0, so x-5 is a factor.
Use long division to get the quadratic factor. As far as real factors are concerned, you are done.
If not, use the quadratic formula to get the other 2 complex roots.

x^3 -125

=x^3 -5^3

= (x-5)(x^2 +5x +25)

a^3 - b^3 ≡ (a - b)(a^2 + ab + b^2)

x^3 - 125
= x^3 - 5^3
= (x - 5)(x^2 + 5x + 25)

x^(3)-125

The binomial can be factored using the difference of cubes formula, because both terms are perfect cubes. The difference of cubes formula is a^(3)-b^(3)=(a-b)(a^(2)+ab+b^(2)).
(x-5)(x^(2)+5x+25)