Thread: algebra

if a+b+c=0, prove (2a-b)^3 +(2b-c)^3 +(2c-a)^3
= 3 (2a-b)(2b-c)(2c-a)

Originally Posted by Frank Chan

if a+b+c=0, prove (2a-b)^3 +(2b-c)^3 +(2c-a)^3
= 3 (2a-b)(2b-c)(2c-a)

Do you remember this formula


x^3 +y^3 +z^3 - 3xyz = (x+y+z)(x^2 +y^2 +z^2 - xy-yz-za)
We will use it

when x+y+z=0

Then x^3+y^3+z^3 = 3xyz
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now in LHS
Consider

2a-b = x

2b -c =y

2c-a= z

Thus x+y+z =
(2a - b) + (2b - c) +(2c -a) = 2(a+b+c) - (a+b+c) = 0

-------------
Hence
x^3 + y^3 +z^3 = (2a-b)^3 +(2b-c)^3 +(2c-a)^3

But using formula

= 3xyz = 3 (2a-b)(2b-c)(2c-a) =RHS

Thus proved

Hello, Frank!

If $\displaystyle a+b+c\:=\:0$, prove: .$\displaystyle (2a-b)^3 +(2b-c)^3 +(2c-a)^3 \:=\: 3 (2a-b)(2b-c)(2c-a)$

Let: .$\displaystyle \begin{array}{cccc}X &=& 2a-b & {\color{blue}[1]} \\ Y &=& 2b-c & {\color{blue}[2]} \\ Z &=& 2c-a & {\color{blue}[3]} \end{array}$

We want to prove that: .$\displaystyle X^3 + Y^3 + Z^3 \:=\:3XYZ$



Add [1], [2], [3]: .$\displaystyle X + Y + Z \:=\:a+b+c \quad\Rightarrow\quad X+Y+Z \:=\:0$ .[4]

. . and we have: .$\displaystyle \begin{array}{ccc} X+Y &=& \text{-}Z \\ X+Z &=& \text{-}Y \\ Y+Z &=& \text{-}X\end{array}$ .[5]


Cube both sides of [4]: .$\displaystyle (X+Y+Z)^3 \:=\:0^3$

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

. . $\displaystyle X^3 + Y^3 + Z^3 + 3XY(X+Y) + 3YZ(Y+Z) + 3XZ(X+Z) + 6XYZ \;=\;0$


Substitute [5]: .$\displaystyle X^3 + Y^3 + Z^3 + 3XY(\text{-}Z) + 3YZ(\text{-}X) + 3XZ(\text{-}Y) + 6XYZ \;=\;0$

. . . . . . . . $\displaystyle X^3 + Y^3 + Z^3 - 3XYZ - 3XYZ - 3XYZ + 6XYZ \;=\;0$

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


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

Thank you for your help.