Giải:
Ta có:
\(B=a^x.b^y\)
\(\Rightarrow B^2=a^{2x}.b^{2y}\)
\(\Rightarrow\left(2x+1\right)\left(xy+1\right)=15\)
\(\Rightarrow15=3.5\Rightarrow\left\{\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Lại có:
\(B^3=a^{3x}.b^{3y}\)
\(\Rightarrow\left(3x+1\right)\left(3y+1\right)\)
\(=\left(3.1+1\right)\left(3.2+1\right)\)
\(=4.7=28\)
Vậy \(B^3\) có \(28\) ước