x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)
x3 - 3x2 + x - 3 = (x2 +1)( x - 3)
x2(x3 + 1) is the best you can do there.
If you mean "factors", the two monomials have the common factor x2. Divide each factor by x2 to get the other factor.
x3 + 3x2 - 6x - 8 = (x - 2)(x2 + 5x + 4) = (x - 2)(x + 1)(x + 4)
if: f(x) = x3 - 4xe-2x Then: f'(x) = 3x2 - [ 4e-2x + 2(4x / -2x) ] = 3x2 - 4e-2x + 4
x3-343
Since x3 is a factor of x5, it is automatically the GCF.
Greatest common factor of x4 and x3 is x3.
The greatest common factor of 1-x3 is 1, dummy.
y5(x3 - 8)
X3 X(X2) X2(X) and, X * X * X
x3+27y3 = (x+3y) · (x2-3xy+9y2)
Choose the lowest of the exponents. The GCF of x3 and x5 is x3
x2(x - 8)
Answer: x (x2-x)
x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)