Factor 4AB^2 - 4A^2B - B^3.

Step-by-step solution

1. 1

   Pull out the common variable factor. Every term contains at least one $B$ (the powers are $2,1,3$), while $A$ is missing from the last term, so $B$ alone is the common factor. The coefficients $4,-4,-1$ share no numeric factor beyond $1$:

   $4AB^2-4A^2B-B^3=B\left(4AB-4A^2-B^2\right)$

2. 2

   Reorder the bracket by descending powers of A. Writing it as $-4A^2+4AB-B^2$ makes the structure visible — the leading coefficient is negative, which is a hint that a minus sign wants to come out.

3. 3

   Factor out the minus sign.

   $4AB-4A^2-B^2=-\left(4A^2-4AB+B^2\right)$

   All three signs inside flip. Skipping this step is what usually blocks people from seeing the square.

4. 4

   Recognise the perfect square. For $4A^2-4AB+B^2$: the outer terms are $(2A)^2$ and $B^2$, and the middle term is $-2\cdot(2A)\cdot B=-4AB$ ✓. So

   $4A^2-4AB+B^2=(2A-B)^2$

5. 5

   Reassemble the factorisation.

   $4AB^2-4A^2B-B^3=B\cdot\left[-(2A-B)^2\right]=-B\left(2A-B\right)^2$

   Since squaring kills the sign, $(2A-B)^2=(B-2A)^2$, so $-B(B-2A)^2$ is the same answer written differently.

6. 6

   Note what the form tells you. Because $(2A-B)^2\ge0$ always, the whole expression has the opposite sign to $B$: negative when $B>0$, positive when $B<0$, and zero when $B=0$ or $B=2A$.

7. 7

   Verify numerically. Comparing the original with $-B(2A-B)^2$ at $60$ random pairs $(A,B)$ drawn from $[-5,5]$ gives agreement to machine precision at every point ✓.