## Precalculus (10th Edition)

$189x^5.$
According to the Binomial Theorem, the nth term in the expansion of $(x+a)^k$ is given by: ${k\choose k-n+1}a^{n-1}x^{k-n+1}.$ Hence here for $n=3,k=7$: ${7\choose 7-3+1}(-3)^{3-1}x^{7-3+1}={7\choose 5}(-3)^{2}x^{5}=189x^5.$