Polynomial expansion with binomials

Expansion. $(1+x)^3=\sum_i\binom{3}{i}x^i$, $(1-x)^7=\sum_j\binom{7}{j}(-1)^jx^j$.

Coefficient of $x^5$. $\sum_{i+j=5}\binom{3}{i}\binom{7}{j}(-1)^j$.

Compute.
- $i=0,j=5$: $1\cdot21\cdot(-1)=-21$
- $i=1,j=4$: $3\cdot35=105$
- $i=2,j=3$: $3\cdot35\cdot(-1)=-105$
- $i=3,j=2$: $1\cdot21=21$

Sum: $-21+105-105+21=0$.

Conclusion. The coefficient of $x^5$ is $0$.