<s>[INST] Factoring a long expression in the form $(a+b)^3 + (c - b)^3 - (c+b)^3$ 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. 
 [/INST] 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 $k=3$.  
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