Definition. The support of a monomial $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ is defined to be the set $\operatorname{supp}(x_1^{\alpha_1}\cdots x_n^{\alpha_n})=\{x_i: \alpha_i >0\}$.
Let $R=k[x_1,...,x_n]$ be a polynomial ring and $I$ be any monomial ideal of $R$. For any monomial $m$ with the property that the support of $m$ is disjoint from the support of any generator of $I$ we have ${\rm pd}(mI)={\rm pd}(I)$, where ${\rm pd}(I)$ is the projective dimension of $I$.
It's more convenient to work with $\operatorname{depth}$ instead of $\operatorname{pd}$. Then we have to prove that $\operatorname{depth}R/mI=\operatorname{depth}R/I$. Use the following ses $$0\to R/(m)\cap I\to R/(m)\oplus R/I\to R/(m,I)\to 0.$$
It's easy to see that $m$ is a non-zero divisor on $R/I$, so $\operatorname{depth}R/(m,I)=\operatorname{depth}R/I-1$. We also have $\operatorname{depth} R/(m)\oplus R/I=\min(\operatorname{depth}R/(m),\operatorname{depth}R/I)=\operatorname{depth}R/I$. Now simply use the depth's lemma and find $\operatorname{depth}R/(m)\cap I=\operatorname{depth}R/I$. But $(m)\cap I=mI$ and we are done.