How to show that $\frac{(n+1)^3 + 6(n+1)^2 + 11(n+1)}{3}=\frac{n^3 + 6n^2 + 11n}{3} + (n+2)(n+3) $

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I was trying to do some induction proofs but got stuck when I needed to show that $$\frac{(n+1)^3 + 6(n+1)^2 + 11(n+1)}{3}$$ and $$\frac{n^3 + 6n^2 + 11n}{3} + ((n+2)(n+3)) $$ are in fact equal. Quite new to this stuff and I could not find a simple way to show it. Do I need to expand both expressions? I would prefer not to do it that way. I would get so many terms that I need to keep track of...

How do you rewrite

$$\frac{n^3 + 6n^2 + 11n}{3} + ((n+2)(n+3)) $$ as $$\frac{(n+1)^3 + 6(n+1)^2 + 11(n+1)}{3}$$

in a simple way?

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you can write the first term in the form $$\frac{(n+1)\left((n+1)^2+6(n+1)+11\right)}{3}=\frac{(n+1)(n^2+8n+18)}{3}$$ factorizing the second term you will get the same as above