# binomial raised to the 3rd power (in reverse)

Given the binomial raising to the 3rd power forumula: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3,$ how can we reverse the action? I mean from the result produce the original binomial. For example how to translate this expression : $2x^3 +18x^2 -3x -27$ to the original $(a+b)^3$ form? I know how to find the $b$ variable but I'm not sure about the $a$.

• $2x^3 +18x^2 -3x -27$ is not the third power of any polynomial. It factors as $(x+9) (2 x^2-3)$. Commented Nov 9, 2015 at 15:57
• you should look at $$(ax+b)^3$$ to make it easier for you to determine the conditions on $a,b$ if they indeed exist. Commented Nov 9, 2015 at 15:58
• @HenningMakholm can you show me the steps how you factored?
– BsD
Commented Nov 9, 2015 at 16:06
• $2x^3+18x^2-3x-27=2x^2(x+9)-3(x+9)=(x+9)(2x^2-3).$ Commented Nov 9, 2015 at 16:07
• Only polyinomials with triple roots can be factored thusly. The vast majority of cubics do not have triple roots and can't be so factored. Commented Nov 9, 2015 at 17:25

As chinny84 suggested:

$(ax +b)^3 = a^3x + 3a^2bx^2 + 3ab^2x + b^3$

So to reduce $2x^3 +18x^2 -3x -27$ to $(ax + b)^3$ or $c(ax + b)^3$ we must solve the following four equations

$a^3 = 2$

$3a^2b = 18$

$3ab^2 = -3$

$b^3 = -27$

(or multiples of 1/c of these equations.) These clearly have no solutions (rational or otherwise) so it can't be factored as $(ax + b)^3$. This shouldn't surprise us as only cubics with a triple root can be so factored.