Partitions of $n$ with exactly 3 parts

Hint: The following can be found somewhat more detailed in this answer.

A generating function for the number of partitions with exactly three parts is \begin{align*} &\frac{1}{(1-x)(1-x^2)(1-x^3)}-\frac{1}{(1-x)(1-x^2)}\\ &\qquad=\frac{1}{(1-x)(1-x^2)}\left(\frac{1}{1-x^3}-1\right)\\ &\qquad=\frac{x^3}{(1-x)(1-x^2)(1-x^3)}\\ &\qquad=x^3+x^4+2x^5+3x^6+4x^7+\color{blue}{5}x^8+7x^9\cdots \end{align*}

Example: There are $\color{blue}{5}$ partitions of $8$ with three summands \begin{align*} 8&=1+1+6\\ &=1+2+5\\ &=1+3+4\\ &=2+2+4\\ &=2+3+3 \end{align*}