Question

Pls help me factorize.Thanks!

Answer 1

We can change the sign of the second term, so that the two parenthesis are the same:

$6(a-1)b^3+3(1-a)b^2 = 6(a-1)b^3-3(a-1)b^2$

Now, both terms have in common a 3, becase the numeric factors are 3 and 6. If we factor the 3, we have

$6(a-1)b^3-3(a-1)b^2 = 3(2(a-1)b^3-(a-1)b^2)$

The parenthesis (a-1) is also in common, so we can factor it as well:

$3(2(a-1)b^3-(a-1)b^2) = 3(a-1)(2b^3-b^2)$

Finally, both terms in the parenthesis contain $b^2$, because it's the power of b with the lowest exponent:

$3(a-1)(2b^3-b^2) =3(a-1)b^2(2b-1)$

So, the factorization is

$3b^2(a-1)(2b-1)$

You can swap the order of the four factors (3, b^2, a-1 and 2b-1) as you like: the multiplication is commutative.