# Math Help - factorisation

1. ## factorisation

Hi, I am attempting the following question, but am stuck:

If $a+b+c=0$ , show that $(2a-b)^3+(2b-c)^3+(2c-a)^3 = 3(2a-b)(2b-c)(2c-a)$

Here is what i have tired:
$(2a-b)^3=8a^3-12a^2b+6ab^2-b^3$
$(2b-c)^3=8b^3-12b^2c+6bc^2-c^3$
$(2c-a)^3=8c^3-12c^2a+6ca^2-a^3$

Adding all those together i get:

$=8(a^3+b^3+c^3)-12(a^2b+b^2c+c^2a)+6(ab^2+bc^2+ca^2)-(a^3+b^3+c^3)$
$=7(a^3+b^3+c^3)-12(a^2b+b^2c+c^2a)+6(ab^2+bc^2+ca^2)$

And then i don't know what to do...

2. Originally Posted by M.R
Hi, I am attempting the following question, but am stuck:

If $a+b+c=0$ , show that $(2a-b)^3+(2b-c)^3+(2c-a)^3 = 3(2a-b)(2b-c)(2c-a)$

Here is what i have tired:
$(2a-b)^3=8a^3-12a^2b+6ab^2-b^3$
$(2b-c)^3=8b^3-12b^2c+6bc^2-c^3$
$(2c-a)^3=8c^3-12c^2a+6ca^2-a^3$

Adding all those together i get:

$=8(a^3+b^3+c^3)-12(a^2b+b^2c+c^2a)+6(ab^2+bc^2+ca^2)-(a^3+b^3+c^3)$
$=7(a^3+b^3+c^3)-12(a^2b+b^2c+c^2a)+6(ab^2+bc^2+ca^2)$

And then i don't know what to do...
That's a good start. Now do the same thing with the right-hand side of the identity. You should find that

$3(2a-b)(2b-c)(2c-a) = 21abc -12(a^2b+b^2c+c^2a)+6(ab^2+bc^2+ca^2)$.

Compare that with the expression for the left-hand side, and you see that what you need to prove is that $a^3+b^3+c^3 = 3abc$. To do that, check that $(a+b+c)^3 = a^3+b^3+c^3 +6abc + 3(a^2b+b^2c+c^2a + ab^2+bc^2+ca^2)$ and $(a+b+c)(bc+ca+ab) = 3abc + (a^2b+b^2c+c^2a + ab^2+bc^2+ca^2)$. Then use the given fact that $a+b+c=0$.

3. Hi Opalg,

Thanks for taking the time to reply. I have done as you advised. But i still don't get it. This is what i tired to do:

$(a+b+c)(bc+ca+ab) = 3abc + (a^2b+b^2c+c^2a + ab^2+bc^2+ca^2)
$

Therefore:
$2 (a+b+c)(bc+ca+ab) = 6abc + 2(a^2b+b^2c+c^2a + ab^2+bc^2+ca^2)$
$(a+b+c)^3 = a^3+b^3+c^3 +6abc + 2(a^2b+b^2c+c^2a + ab^2+bc^2+ca^2) +$ $(a^2b+b^2c+c^2a + ab^2+bc^2+ca^2)$
$(a+b+c)^3 = a^3+b^3+c^3 + 2 (a+b+c)(bc+ca+ab) + (a^2b+b^2c+c^2a + ab^2+bc^2+ca^2)$

But:
$(a^2b+b^2c+c^2a + ab^2+bc^2+ca^2) = (a+b+c)(bc+ca+ab) - 3abc
$

Therefore:
$
(a+b+c)^3 = a^3+b^3+c^3 + 2 (a+b+c)(bc+ca+ab) + (a+b+c)(bc+ca+ab) - 3abc
$

Now using the fact $a+b+c=0$ :

$(0)^3 = a^3+b^3+c^3 + 2(0)(bc+ca+ab) + (0)(bc+ca+ab) - 3abc$
$0 = a^3+b^3+c^3 - 3abc$
Hence: $a^3+b^3+c^3 = 3abc$

This solution seems very complicated for a Chapter 1, Year 11 book? And how did you think of this equation $(a+b+c)(bc+ca+ab) = 3abc + (a^2b+b^2c+c^2a + ab^2+bc^2+ca^2)$ ???

4. Hello, M.R!

If $a+b+c\:=\:0$, show that: . $(2a-b)^3+(2b-c)^3+(2c-a)^3 \:=\: 3(2a-b)(2b-c)(2c-a)$
Let: . $\begin{Bmatrix}u \:=\:2a-b \\ v \:=\:2b-c \\ w \:=\:2c-a\end{Bmatrix}$

Note that: . $u+v+w \:=\:(2a-b) + (2b-c) + (2c-a) \:=\:a+b+c$
. . .Hence: . $u+v+w \:=\:0$ .[1]

We have: . $u^3 + v^3 + w^3 \:=\:3uvw$ . . . which can be proved.

Cube [1]: . $(u+v+w)^3 \:=\:0^3$

. . . . $u^3 + v^3 + w^3 + 3u^2v + 3uv^2 + 3v^2w+3vw^2 + 3u^2w + 3uw^2 + 6uvw \;=\;0$

. . . . $u^3 + v^3 + w^3 + (3u^2v + 3uv^2 + 3uvw) + (3v^2w + 3vw^2 + 3uvw) + 3u^2w + 3uw^2 \;=\;0$

. . . . $u^3 + v^3 + w^3 + 3uv\underbrace{(u+v+w)}_{\text{This is 0}} + 3vw\underbrace{(u+v+w)}_{\text{This is 0}} + 3uw\underbrace{(u+w)}_{\text{This is }-v} \;=\;0$

. . . . $u^3 + v^3 + w^3 - 3uvw \;=\;0$

Therefore: . $u^3 + v^3 + w^3 \;=\;3uvw\quad\hdots$ .QED!

5. Hi Soroban,

Firstly thank you for taking the time to reply. You provided me with an excellent solution, and i understand majority of it. I just can't seem to figure out one line, and that is:

Originally Posted by Soroban
Note that: . $u+v+w \:=\2a-b) + (2b-c) + (2c-a) \:=\:a+b+c" alt="u+v+w \:=\2a-b) + (2b-c) + (2c-a) \:=\:a+b+c" />
How does $(2a-b) + (2b-c) + (2c-a) \:=\:a+b+c$ ?

If i go by the substitution: $\begin{Bmatrix}u \:=\:2a-b \\ v \:=\:2b-c \\ w \:=\:2c-a\end{Bmatrix}$

Then $a = 2c -w \:, b = 2a - u \: and \:c = 2b - v$ ? I just don't get, how you equated the two function to make $u + v + w = a + b + c$.

From that definition we can also deduce: $a = 2a-b \:, b=2b-c \: and \: c=2c-a$. Hence $a=b \:, b=c \: and \: c=a$. So $a=b=c$??? And if that's the case, then only 0 and 1 would satisfy that equation? Is this true?

6. Lemma: $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\left( a^{2}+b^{2}+c^{2}-ab-bc-ac \right).$

Proof: just write $a^{3}+b^{3}+c^{3}-3abc=(a+b)^{3}+c^{3}-3ab(a+b)-3abc,$ and factorise.

As for the problem, let's rewrite the LHS as $(2a-b)^{3}+(2b-c)^{3}+(2c-a)^{3}-3(2a-b)(2b-c)(2c-a)+3(2a-b)(2b-c)(2c-a),$ thus applying the lemma we only take care about the first factor, which is $\big(2(a+b+c)-(a+b+c)\big)=0,$ and the only thing what remains is $3(2a-b)(2b-c)(2c-a),$ and we're done.

7. Thanks all for you help. I appreciate your help.

For one reason or another, the solution is just not getting to me. Maybe i will try it in a weeks time and just leave it for now. Spent way too much time on one question. Will continue and get back soon Thanks to all again

8. Hello M.R.
Originally Posted by M.R
...
How does $(2a-b) + (2b-c) + (2c-a) \:=\:a+b+c$ ?
As you said, you have been too long on this question. Look at what you have just written here. Remove the brackets. Not very hard, is it?

9. Originally Posted by M.R
Thanks all for you help. I appreciate your help.

For one reason or another, the solution is just not getting to me. Maybe i will try it in a weeks time and just leave it for now. Spent way too much time on one question. Will continue and get back soon Thanks to all again
I really suggest you to take a harder look at the solutions, specially on Soroban's one, he has an excellent teaching and he spends a lot of time here so that people understand how to solve the problems.

My solution is very simple too, so read it carefully. - It's just simple algebra, and there's no much here.