Factorise the following expression: $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$

(N/A) The given expression is $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$.
We can rewrite this expression using the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}$.
Here,let $a = 2x$ and $b = 3y$.
Then,$a^{3} = (2x)^{3} = 8x^{3}$ and $b^{3} = (3y)^{3} = 27y^{3}$.
The middle terms are $3a^{2}b = 3(2x)^{2}(3y) = 3(4x^{2})(3y) = 36x^{2}y$ and $3ab^{2} = 3(2x)(3y)^{2} = 3(2x)(9y^{2}) = 54xy^{2}$.
Thus,the expression is of the form $(2x)^{3} + (3y)^{3} + 3(2x)^{2}(3y) + 3(2x)(3y)^{2}$.
This simplifies to $(2x + 3y)^{3}$.
Therefore,the factorised form is $(2x + 3y)(2x + 3y)(2x + 3y)$.