1. ## Challenging factorization question?

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

2. ## Re: Challenging factorization question?

Originally Posted by magakriv
If a+b+c = 0 show that (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)
Use this webpage to help do the algebra.

3. ## Re: Challenging factorization question?

Hello, magakriv!

$\displaystyle \text{If }a+b+c \,=\,0,$
. . $\displaystyle \,\text{ show that: }\:(2a-b)^3 + (2b-c)^3 + (2c-a)^3 \:=\:3(2a-b)(2b-c)(2c-a)$
Let: .$\displaystyle \begin{Bmatrix}X &=& 2a-b \\ Y &=& 2b -c \\ Z &=& 2c-a\end{Bmatrix}$

Note that: /$\displaystyle X + Y+Z \:=\:a+b+c \:=\:0$

We have: .$\displaystyle X+Y+Z \;=\;0$

Cube both sides: .$\displaystyle (X+Y+Z)^3 \;=\;0^3$

. . $\displaystyle X^3+Y^3+Z^3+3X^2Y+3XY^2+3Y^2Z+3YZ^2+3Z^2X+3ZX^2+6X YZ \;=\;0$

$\displaystyle X^3+Y^3+Z^3+(3X^2Y+3XY^2+3XY\!Z)+(3Y^2\!Z+3Y\!Z^2+ 3XY\!Z)+$$\displaystyle (3X^2Z+3XZ^2) \;=\;0$

$\displaystyle X^3+Y^3+Z^3 + 3XY(X+Y+Z) + 3YZ(Y+Z+X) + 3XZ(X+Z) \;=\;0$

Since $\displaystyle X+Y+Z \,=\,0\,\text{ and }\,X+Z \,=\,-Y$, we have:

. . $\displaystyle X^3+Y^3+Z^3 + 3XY(0) + 3Y(0) + 3XZ(-Y) \;=\;0$

. . . . . . $\displaystyle X^3+Y^3+Z^3 - 3XYZ \;=\;0$

. . . . . . $\displaystyle X^3 + Y^3 + Z^3 \;=\;3XYZ$

Back-substitute: .$\displaystyle (2a-b)^3 + (2b-c)^3 + (2c-a)^3 \;=\;3(2a-b)(2b-c)(2c-a)$

4. ## Re: Challenging factorization question?

Originally Posted by magakriv
If a+b+c = 0 show that (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)
$\displaystyle \text{Let's first consider the cubic expansion of three variables.}$

\displaystyle \begin{aligned} (x+y+z)^3 &= x^3 +y^3 +z^3 +3x^2y+3x^2z + 3y^2x + 3y^2z +3z^2x + 3z^2y + 6xyz\\ &=x^3 + y^3 + z^3 + (3x^2y+3y^2x+3xyz) + (3z^2y+3y^2z+3xyz) + (3z^2x+3x^2z+3xyz) -3xyz\\ &=x^3 + y^3 + z^3 +3xy(x+y+z) +3yz(x+y+z) +3zx(x+y+z) - 3xyz \end{aligned}

$\displaystyle \text{Now let return to our problem. Given}~ a+b+c=0.}$

$\displaystyle \text{Show that}~ (2a-b)^3 + (2b-c)^3 + (2c-a)^3=3(2a-b)(2b-c)(2c-a)$

$\displaystyle \text{Let}~ x=2a-b,~ y=2b-c,~ z=2c-a.~ \text{Observe that}~ x+y+z=a+b+c=0$

$\displaystyle \text{Substitute}~ x+y+z=0~ \text{into the derivation above, we have}$

$\displaystyle 0=x^3+y^3+z^3 - 3xyz~ \Leftrightarrow~x^3+y^3+z^3 = 3xyz$

$\displaystyle \text{Thus,}~ (2a-b)^3 + (2b-c)^3 + (2c-a)^3=3(2a-b)(2b-c)(2c-a)$

Let me know if you have any question. Good luck!