Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to factor the following:$$\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \left(\dfrac{3}{4}z -\dfrac{5}{3}y\right)^3 - \left(\dfrac{3}{4}z + \dfrac{2}{3}x\right)^3$$A friend of mine suggested that $(a + b)^3 +(c - b)^3 - (c + b)^3 = 3(a+b)(a+c)(b +a)$, and it's right. But this is just a normal exam question in 9th grade... so it must have a “normal” way to be done. I tried factoring $(a+b)^3 + (c - b)^3$ by using $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ but all hell breaks loose.

share|cite|improve this question
factorise the last two terms first. – Easy Apr 14 '13 at 5:32
Internet problems I have so cannot come on chat. Heed the first comment. – Jayesh Badwaik Apr 14 '13 at 5:40
$(c - b)^3 - (a + b)^3 = \left(c - a - 2b\right)\left(c^2 - 2cb + ac - ba + bc + a^2 + 2ab + b^2 \right)$ Is that correct? – Parth Kohli Apr 14 '13 at 5:40
@pen you have $(c-b)^3 - (c+a)^3$. – Jayesh Badwaik Apr 14 '13 at 5:42
@pen, then you can taking out the common factor with the first term and fatorise the rest. – Easy Apr 14 '13 at 5:53
up vote 2 down vote accepted

You have $\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \left(\dfrac{3}{4}z -\dfrac{5}{3}y\right)^3 - \left(\dfrac{3}{4}z + \dfrac{2}{3}x\right)^3$, which (as you observed) is of the form $(a+b)^3 + (c-b)^3 - (c+a)^3$.

A short cut is to simply observe that if $a = -b$ or $-c$, the expression is $0$. Similarly, you should be able to easily observe that $b=c$ also leads to the expression being $0$. This gives you $(a+b)(a+c)(b-c)$ as factors immediately. Hence

$(a+b)^3 + (c-b)^3 - (c+a)^3 = k(a+b)(a+c)(b-c)$, for some scalar $k$.
To find $k$, you could compare coefficients of some power or test a suitable value. For e.g. let $a = 0, b = 1, c =-1$. Then LHS $= 1 - 8 + 1 = -6,$ and RHS $= -2k$, so $k=3$.

share|cite|improve this answer
Nice alternative. Thanks. – Parth Kohli Apr 14 '13 at 9:47

Perhaps another way to see this is to carry out the binomial-cubes using the "Pascal coefficients" and see what the sum leaves:

$$(a + b)^3 + (c - b)^3 - (c + a)^3 =$$

$$(a^3 + 3a^2b + 3ab^2 + b^3)$$ $$+(c^3 - 3c^2b + 3cb^2 - b^3)$$ $$-(c^3 + 3c^2a + 3ca^2 + a^3)$$

$$3 (a^2b + ab^2 - c^2b + cb^2 - c^2a - ca^2),$$

after cancelling all the cubic terms. The remaining six terms contain every possible group of the three factors $a, b,$ and $c$ in which two of these repeat. So we have a product of factors $3 (a \pm b)(a \pm c)(b \pm c)$, covering the various combinations.

The negative terms all contain a factor of $c$ , but the term with $cb^2$ is positive. That and the other positive terms can be formed from the product $3 (a + b)(a + c)(b \pm c)$. The only way we could have the terms $-c^2a$ and $-c^2b$ is for the two appearances of $c$ to carry opposite signs; this will also get us the term $-ca^2$. So the factors must be $3 (a + b)(a + c)(b - c)$.

But wait: shouldn't this product have eight terms? It does, and they are $a \cdot c \cdot b$ and $b \cdot a \cdot [-c]$. So they cancel, which also explains the single negative term $-ca^2$, with just one factor of $c$.

Not the quickest method of solving this, but possibly suggesting another way to view such factorizations.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.