Simplify the fraction with numerator L squared minus R squared and denominator H plus R.

Step-by-step solution

1. 1

   Factor the numerator with the difference of squares. Since $L^2-R^2$ is $a^2-b^2$ with $a=L$ and $b=R$,

   $L^2-R^2=(L-R)(L+R)$

   so the fraction becomes $\dfrac{(L-R)(L+R)}{H+R}$. Factoring is always the right first move: cancellation is only ever possible between factors, never between individual terms.

2. 2

   Compare the available factors with the denominator. The numerator offers $(L-R)$ and $(L+R)$; the denominator is $(H+R)$. These are not the same expression — $(L+R)$ and $(H+R)$ agree in the $R$ but differ in the leading letter, and there is no rule that lets an $L$ cancel an $H$.

3. 3

   State the simplified form. With no common factor, the work stops here:

   $\frac{L^2-R^2}{H+R}=\frac{(L-R)(L+R)}{H+R}$

   This is "simplified" in the sense that the numerator is fully factored; the fraction itself cannot be reduced.

4. 4

   Identify the one special case. If the problem additionally tells you $H=L$, the denominator becomes $L+R$, which does match a factor:

   $\frac{(L-R)(L+R)}{L+R}=L-R\qquad (L+R\neq 0)$

5. 5

   Demonstrate with numbers why the general cancellation is wrong. Take $L=5$, $R=3$, $H=2$: the fraction is $\frac{25-9}{2+3}=\frac{16}{5}=3.2$, while $L-R=2$. They differ, so cancelling would have been an error. With $H=L=5$ instead: $\frac{16}{8}=2=L-R$ ✓, exactly as the special case predicts.