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 , prove: .

Let: .

We want to prove that: .



Add [1], [2], [3]: . .[4]

. . and we have: . .[5]


Cube both sides of [4]: .

. .

. .


Substitute [5]: .

. . . . . . . .

. . . . . . . .


Therefore:.

Thank you for your help.