150. $N=p^{\lambda_{1}} q^{\lambda_{2}}$, where $p$ and $q$ are different prime numbers; $N^{2}$ has 15 divisors. How many divisors does the number $N^{3}$ have?
150. Since $N^{2}=p^{2 \lambda_{1}} q^{2 \lambda_{2}}$, then $\left(2 \lambda_{1}+1\right)\left(2 \lambda_{2}+1\right)=15$, from which $\left\{\begin{array}{l}2 \lambda_{1}+1=3 \\ 2 \lambda_{2}+1=5\end{array}, \lambda_{1}=1, \lambda_{2}=2\right.$.

Given that $N^{3}=p^{3 \lambda_{1}} q^{3 \lambda_{2}}$, the number of all divisors of the number $N^{3}$ is $\left(1+3 \lambda_{1}\right)\left(1+3 \lambda_{2}\right)=28$.