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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.

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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
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@pen, then you can taking out the common factor with the first term and fatorise the rest. –  Easy Apr 14 '13 at 5:53

2 Answers 2

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 $k=3$.

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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.

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