A nice question

2)
WLOG,a>b>c

so, a-b>0
b-c>0
c-a<0

that implies, (c-a)^1/3 is imaginary whereas other 2 expressions are real.

so,sum of real and imaginary number can never be zero(a,b,c distinct)

nice explanation....

both of them are based on single concept

Adding one more.

prove the AM-GM inequality for 3 terms ie x³ + y³ + z³ ≥3xyz

(-8)^1/3=(8)^1/3.(-1)^1/3

= 2.(-1) or 2.(ω) or 2.(ω²)

quesn 1 wud be solved by the binomial method
:)

let ³√(a-b)=x,³√(b-c)=y,³√(c-a)=z.

then,x³+y³+z³=0

either,(x+y+z)=0
or,x=y=z...[then,it will be a=b=c]
oh...my god..i have proved..
³√(a-b)+³√(b-c)+³√(c-a)=0

@ #10 you have a³ + b³ = 4 (not zero)

so (a+b)³ - 3ab(a+b) = 4

or (a+b)³ + 3(a+b) = 4 (ab= -1)

solving the equation for (a+b) we will get (a+b) = 1 hence its proved that the expression is rational.

it can also be solved this way

let x= (2+√5)^1/3 + (2-√5)^1/3

so x- (2+√5)^1/3 - (2-√5)^1/3=0

so x³ - 2- √5 - 2 + √5= 3x (2+√5)^1/3 (2-√5)^1/3

(using the concept that if a+b+c = 0, a³ + b³ +c³ =3abc)

or x³ - 4 = - 3x

or x=1

there is no other real root.

so the expression is a rational number.

http://www.targetiit.com/iit-jee-forum/posts/area-volume-18431.html

@Soham -> Thanks for clearing my doubt

Well your solution was wonderful and easy to understand....!!

Thanks..!!