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just for safety:
a\(\ne\)b
a\(\ne\)-b

hmm....nice try
no other formula for a^3+b^3 ?

yes, question is correct.

how ?

continuing from parthkohli, |dw:1347876944383:dw|

and your question simplifies to |dw:1347876996875:dw|

neat solution

yes, thats right, its 0.there is easier way. rather than continuing...

\((a^3+b^3)=(a+b)^3-3ab(a+b)=0\)
and the rest is ............

haha... yeah :D

well even though we solved it, it IS technically wrong I think? :\

|dw:1347877692079:dw|

|dw:1347877816341:dw|

if i read it correctly, there exist no a,b not equal to 0, such that a^2+b^2-ab = 0.....??

Yea...to be clearer, I mean that a^2+b^2-ab=0 iff a=b=0

u sure about a^2+b^2-ab=0 iff a=b=0 ?

algebraically, x^3, is a one one function... so,
a^3 +b^3=0
implies a^3=(-b)^3
which implies a=-b

a^3 =-b^3 have 3 roots.....i think

Graphically,

If your question covers complex numbers, then it's fine.... but for real, this is it :)

@hartnn Thanks for the time man.. that was a good exercise for me ^_^

you're most welcome buddy!