Ask Uncle Colin: Some Symmetric Algebra

Dear Uncle Colin,

If I know that $a+b+c = 0$, how can I show that $(2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a)$?

- Something You Might Merrily Explain? Thanks! Regards! Yippee!

Hi, SYMMETRY, and thanks for your message!

As usual, there are myriad ways to attack this, of which I can see two straight away: we can eliminate a variable, or we can do some clever cubing.

Eliminating a variable

For example, we could use the first equation to state that $c = -(a+b)$, making the result we want to show equivalent to $(2a-b)^3 + (3b+a)^3 - (3a + 2b)^3 = -3(2a-b)(3b+a)(3a+2b)$.

From there, we could carefully multiply out each of the cubics, do the same with the right hand side and say “ta-da! they’re the same.” But we’re not going to do that.

Instead, we can use the difference of cubes formula, $m^3 -n^3 = (m-n)(m^2 + mn + n^2)$, to say that the last two terms on the left are $(b-2a)\br{ (3b+a)^2 + (3b+a)(3a+2b) + (3a+2b)^2}$ - which has a common factor with the first term!

So we now have a left hand side of $(2a-b)\left[ (2a-b)^2 - (3b+a)^2 - (3b+a)(3a+2b) - (3a+2b)^2 \right]$.

Expanding the big square bracket gives $ (4a^2 - 4ab + b^2) - (a^2 + 6ab + 9b^2) - (3a^2 + 11ab + 6b^2) - (9a^2 + 12ab + 4b^2)$ at the first iteration; being super-careful with the subtracted brackets, that becomes $-9a^2 - 33ab - 18b^2$. That has a factor of -3, which is promising: taking that out gives $3a^2 + 11ab + 6b^2$, which factorises as $(3a+2b)(a+3b)$ - so the left hand side is equal to $-3(2a-b)(3a+2b)(a+3b)$, which is the target right hand side! $\blacksquare$

But there’s something a bit messy about that: the original set-up has a beautiful symmetry to it, and it seems a shame not to exploit it.

Clever cubing

We know that $a+b+c=0$, so we know that $(a+b+c)^3 = 0$ and - equivalently - $\left[(2a-b) + (2b-c) + (2c-a)\right]^3 = 0$.

I’m going to call those three terms $C$, $A$ and $B$, respectively.

Multiplying out, we get $(A+B+C)^3 = A^3 + B^3 + C^3 + 3A^2B + 3B^2C + 3C^2A + 3 AB^2 + 3BC^2 + 3CA^2 + 6ABC = 0$ (a quick mental check: that’s 27 terms altogether, as expected).

Ignoring the very last term, let’s factor out $A^2$, $B^2$ and $C^2$:

$A^2(A + 3B + 3C) + B^2(B + 3C + 3A) + C^2(C +3A + 3B) + 6ABC = 0$

BUT! $A+B+C=0$, so that’s $-2A^3 - 2B^3 - 2C^3 + 6ABC = 0$, or $A^3 + B^3 + C^3 = 3ABC$.

Replacing our made-up letters: $(2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a)$ $\blacksquare$


I imagine there are even more elegant approaches - but I hope that helps.

- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

Share

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter