Cube root of a binomial

Assume that the degree of $a$ is larger than the degree of $b$ so that its expansion is ordered in descending powers of $y$. Then we can match the highest and lowest degree terms: $$ a^3 = 8y^3 \implies a = 2y \\ b^3 = -27 \implies b = -3 $$ Finally, verify that our guess is correct by expanding $(2y - 3)^3$ and checking that it matches the original polynomial.

Think about what the cube root would look like and how you can algebraically represent its cube. What kind of binomial, when cubed, would yield a cubic in one variable?

Simply look at the last two terms. $$(a-b)^3=a^3-3a^2b+3ab^2-b^3\tag{1}$$ and from the binomial $$8y^3-36y^2+54y-27\tag{2}$$

We see that $$\begin{cases}8y^3=a^3\\27=b^3\end{cases}\tag{3}$$

Solving, we see that $a=2y$ and $b=3$. So the binomial factors into $$(2y-3)^3\tag{4}$$

Expanding out $(4)$ to check, we get: $$8y^3-27-3(2y)^2(3)+3(2y)(9)\tag{5}\\=8y^3-9\cdot4y^2+54y-27\\=8y^3-36y^2+54y-27$$

Which is equal to $(2)$.