# 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...

• If you have $a^3=8, b^3=(-)27$ in $a^3x^3+\dots+(-)27$, what do you think could be the solution? Commented Sep 21, 2016 at 3:09

## 4 Answers

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?

• a bit ambiguous... Commented Sep 21, 2016 at 3:05
• @thunderbolt This better? Commented Sep 21, 2016 at 3:10
• binomial with second term 0 degree? Commented Sep 21, 2016 at 3:13
• @thunderbolt Yes. Now think about what of linear binomial, when cubed, is $8y^3-36y^2+54y-27$. Commented Sep 21, 2016 at 3:14
• Like in the form (ay-b) Commented Sep 21, 2016 at 3:15

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 $(2)$.

We can see that the plus and minus pattern follows the $(a-b)^3$ expansion. Then simply equate coefficients in any two terms:

$8y^3 = a^3 \implies a = 2y$

$27 = b^3 \implies b = 3$

So the cube root is $(2y-3)$