Prerequisites - P.3 - Polynomials and Special Products - Exercises - Page 31: 50

$27x^3+54x^2y+36xy^2+8y^3$

Let's use the concept of cubing a binomial given by $$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$Hence, $\begin{align}(3x+2y)^3&=(3x)^3+3(3x)^2(2y)+3(3x)(2y)^2+(2y)^3\\&=27x^3+3(9x^2)(2y)+3(3x)(4y^2)+8y^3\\&=27x^3+54x^2y+36xy^2+8y^3\end{align}$