Solution
= Solution
The monomials $x^iy^jz^k$ with nonnegative integers $i+j+k=m$ form a <basis> for the <vector space> of homogeneous <polynomials> of degree $m$. For each $i=0,\ldots,m$, there are $m-i+1$ choices of $(j,k)$. Hence
$$
\boxed{\dim V_m=\sum_{i=0}^m(m-i+1)=\binom{m+2}{2}
=\frac{(m+1)(m+2)}2.}
$$