Factor the difference (a + x) to the power m+1 times (b + x) to the power n−1, minus (a + x) to the power m times (b + x) to the power n.

Step-by-step solution

1. 1

   Compare the exponent of each base across the two terms. Do not expand anything — the letters $m$ and $n$ make that impossible. Instead line the powers up:

   $\text{base } (a+x):\quad m+1 \text{ and } m$

   $\text{base } (b+x):\quad n-1 \text{ and } n$

2. 2

   Take the lowest power of each base as the common factor. This is the general GCF rule for powers, and it works exactly the same when the exponents are letters:

   $\gcd=(a+x)^{m}(b+x)^{n-1}$

   The first term keeps one extra $(a+x)$; the second keeps one extra $(b+x)$.

3. 3

   Factor it out and read off what is left. Using $u^{m+1}=u^m\cdot u$ and $v^{n}=v^{n-1}\cdot v$,

   $(a+x)^{m+1}(b+x)^{n-1}-(a+x)^m(b+x)^n=(a+x)^m(b+x)^{n-1}\left[(a+x)-(b+x)\right]$

4. 4

   Simplify the bracket — this is where the $x$ disappears.

   $(a+x)-(b+x)=a+x-b-x=a-b$

   The variable cancels completely, leaving a constant difference of the two parameters.

5. 5

   Write the final factorization and sanity-check it.

   $(a-b)(a+x)^m(b+x)^{n-1}$

   A quick numeric test with $a=3,b=2,x=1,m=2,n=3$: the original is $4^3\cdot 3^2-4^2\cdot 3^3=576-432=144$, and the factored form gives $(3-2)\cdot 4^2\cdot 3^2=144$. They match. Note the result is zero exactly when $a=b$, which makes sense because the two original terms are then identical.