problem solving (1 Viewer)

[byakuya kuchiki]

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

please help.

 

[Timothy.Siu]

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

please help.

i gues u cud expand it lol

edit: u cud also let a=-b-c and sub it into both sides
then, u can factorise two of the terms on LHS using sum of cubes, and then u can factorise the other term using common factor. and then i think u have to expand the other stuff.

 

[h3ll h0und]

byakuya/wang what topic r u doing for maths at apcs?

 

[Drongoski]

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

please help.

Where is this question from? Is this a harder 3U question ?

 

[Iruka]

We solve this problem by reducing it to something more manageable.
Let x=2a-b, y= 2b-c, z=2c-a.

Then x+y+z= 2a-b+2b-c+2c-a = a+b+c = 0, by assumption.
So we have to show that if x+y+z=0, then x^3+y^3+z^3-3xyz=0.

But
(x+y+z)^3 = x^3+y^3+z^3 +3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z) +6xyz
=x^3+y^3+z^3-3xyz +3(3xyz+x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)
= x^3+y^3+z^3-3xyz +3[xy(x+y+z)+ yz(x+y+z)+ xz(x+y+z)]
Since x+y+z=0, LHS=0 and so does all the stuff in the square brackets.

So x^3+y^3+z^3-3xyz=0, as required.

 

[ibrian]

probably polynomials relation btwn roots