<s>[INST] Cube root of a binomial The cube of a certain binomial is $8y^3-36y^2+54y-27$. Find the binomial.
I know that $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and that$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ but don't know how to go further...
 [/INST] 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 $(2)$.
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