Proving Equalities: Check My Work

[armolinasf]

Homework Statement

I posted an inequality proof question a little while ago and got some really great feedback. Here's a couple more that are similar to it that I've been working on. I'm still uncertain as to what makes an acceptable proof and what doesn't, so please let me know if I'm on the right track:


A) Prove that if a+b+c=0 then a^3+b^3+c^a=3abc.

B) Prove that if a^2+ab+b^2=0 then a=0 and b=0.
2. The attempt at a solution

For the first one, my thinking is that since

(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=a^3+b^3+c^a-3abc=0 if a+b+c=0

But this is the same as a^3+b^3+c^a=3abc

For the second, could we just say that since a=0 and b=0 then a=b, so a^2+ab+b^2=0 can become either 3a^2 or 3b^2, and if a=b=0 then 3a^2=3b^2=0

 

[Dick]

The clever factorization works fine for the first one. For the second one, you can't assume a=b. You used the fact you want to prove to derive that. I would suggest you try and complete the square in a on the part of the expression a^2+ab.

 

[armolinasf]

after completing the square I'm getting (a+b/a)^2+(b-b/2)(b+b/2)=0. So if i try and solve for a I get a negative discriminant which would mean that there are no real roots (right?).

So that means since there are no roots a=0 and b=0 for a^2+ab+b^2=0

But something is telling me that that is not quite right...?

 

[Dick]

after completing the square I'm getting (a+b/a)^2+(b-b/2)(b+b/2)=0. So if i try and solve for a I get a negative discriminant which would mean that there are no real roots (right?).

So that means since there are no roots a=0 and b=0 for a^2+ab+b^2=0

But something is telling me that that is not quite right...?

Just leave it as (a+b/a)^2+3b^2/4=0. Now you've written it essentially as the sum of two squares. If x^2+y^2=0 what does that tell you about x and y?

 

[armolinasf]

if x^2+y^2=0 then they must be equal to zero right? Because they can't be opposites of each other since any negative would be canceled by the squaring.

Also, I was wondering if you factor it as a difference of squares:
(a+b/2)^2-(sqrt(-3)b/2)^2 ?

 

[Dick]

if x^2+y^2=0 then they must be equal to zero right? Because they can't be opposites of each other since any negative would be canceled by the squaring.

Also, I was wondering if you factor it as a difference of squares:
(a+b/2)^2-(sqrt(-3)b/2)^2 ?

x^2>=0, y^2>=0. Sure, so x^2+y^2=0 only if x=y=0. Now why would you want to factor as a difference of squares with an imaginary sqrt(-3)?? Leave it as the sum of squares and conclude they both must be zero.