Simplify 9 times (m + n) squared, minus (m - n) squared.

Step-by-step solution

1. 1

   Notice that both terms are perfect squares. Since $9=3^2$,

   $9(m+n)^2=\bigl[3(m+n)\bigr]^2=(3m+3n)^2$

   so the expression is $A^2-B^2$ with $A=3m+3n$ and $B=m-n$. Recognising this avoids expanding two squares and then re-factoring.

2. 2

   Apply $A^2-B^2=(A-B)(A+B)$.

   $(3m+3n-(m-n))\bigl(3m+3n+(m-n)\bigr)=(2m+4n)(4m+2n)$

   The first bracket flips both signs of $B$: $3n-(-n)=4n$.

3. 3

   Take the numeric factors out of each bracket.

   $(2m+4n)(4m+2n)=2(m+2n)\cdot 2(2m+n)=4(m+2n)(2m+n)$

4. 4

   Cross-check by expanding the original.

   $9\left(m^2+2mn+n^2\right)-\left(m^2-2mn+n^2\right)=9m^2+18mn+9n^2-m^2+2mn-n^2$

   which collects to $8m^2+20mn+8n^2$. Note the middle terms add ($18mn+2mn$) because the minus sign flips the $-2mn$.

5. 5

   Confirm the two forms agree.

   $4(m+2n)(2m+n)=4\left(2m^2+5mn+2n^2\right)=8m^2+20mn+8n^2$ ✓

6. 6

   Numerical check. At $m=2$, $n=1$: the original is $9(9)-1=80$; the expansion gives $32+40+8=80$; the factored form gives $4(4)(5)=80$. All three match.