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............. eq(1)
....................... eq(2)
{ b x eq(1) } - { c x eq(2) } gives
As 
is irrational, so .. b2 - ac = 0 and ab = 2c2.
So c422 = a2b2 = a3c. If c 
0, 22 = a3 / b3 which is false as 4 = 22 is not a prefect cube.
So c = 0. And this means a = b = c =0.
Rate if i am correct.
No need for calculation at all !!!
Sum of three irrational no = a ratonal nos ( the irrational no's being such that none of them is expressible as a ratnal linear combination of any other )
Hence they are all zero individually .
Proof : We know that if x and y are two irrational nos ( with the above mentioned condition ) they are individually zero .
Now for three nos x,y,z we apply the above theorem for (x+y) and z ( both irrational )
Hence x+y = 0 , z= 0
hence x=y=z=0
Well, this prob is already there in the 11th class (?) text book by R D Sharma and I think many of you are already of that solution. The reason I posted this problem is because of a solution I saw in a book which illustrated a beautiful method for doing this problem, which will remind you of the way the irrationality of 
Before I start that, lets look at the solution posted above because its a classic example of circular logic. In layman terms - which came first the chicken or the egg. Another metaphor that comes to mind: "putting the cart before the horse"
To explain, suppose I ask evaluate: 
normally, you would say it is equivalent to 
But every once in a while you will meet someone who will drawl: well, we know that this nothing but the derivative of f(x) = x2 at x = 2 and hence the limit is 4. We will then have to point out that you are putting the cart before the horse, because actually, that derivative is arrived at by calculating this limit in the way given above
Likewise, when you say "( the irrational no's being such that none of them is expressible as a ratnal (sic) linear combination of any other ) " someone has to point out that that is precisely the object of the problem. The result of the problem itself has been used to prove the problem! And that deserves 
Ok, now for the serious stuff:
![a+ b \sqrt [3] 2 + c \sqrt [3] 4 = 0](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/b/5/0/b501a38854a476b5836c2717fe3cce7cc45828da.gif)
We can actually assume that a,b and c are integers for if they were not, by multipying out by the LCM, you can convert it into an equation in integers. Also, then it follows that not all of a,b and c are even for if they would be, we can go on dividing the equation by 2 till we reach a stage when atleast one of them is odd. i.e. gcd(a,b,c) > 2
(using a well known identity)
You can easily see that a has to be even. So, let 
The equation becomes 
Divide by 2 and you get,
In the same way, you can now set, 
for the next iteration and similarly next, 
This means that a, b and c are all even contradicting our original statement that atleast one of a, b and c is odd.
Finito!
The above is a classic example of mis understanding ( or unability to understand) a subtle logic !!(it also deserves the same )
I think The above expert has forgotten the basic stuff about mathematics and also a li'l bit of english as he has completely missed the point !!! ( don't mind ,it happens at this age !!!)
the irrational no.s that I was talking about was 
and
.
My point was is not expressible as = q ( for some rational q ) ( it is obvious as then q= ( 1/ !!!!)
Then I proved a theorem : It says that if bx +cy = 0 ( where b,c are rationals and x,y are irrational of the type mentioned ) then b=c=0
This follows obviously as if c!=0 then y =(-b/c)x thus violating the condition . Hence c= 0 similarly b=0
Similarly consider a +bx +cy = 0 ( as is the case here )
so a rational no + an irrational no = 0
so both are zero .( otherwise a ratonal can't be an irrational one )
i.e.a=b=c=0
Sorry, age makes me dense at times. I could follow your argument right up to where you deftly proved that bx+yc cannot be zero for non-zero b and c.
However, in the next step, it has been directly assumed without justification that bx+cy is an irrational number. (Quoting from the above post: so a rational no + an irrational no = 0)
Kindly prove that bx+cy is not equal to any rational number. It is not enough to prove that it cannot be equal to zero, as zero is just one of infinitely many rational numbers.
And that is precisely the statement of the problem. And the basis for my statement that the method presented by you constitutes what is known as circular logic.
This is solved by Titu Andreescu, who trains the USA team for the International Mathematics Olympiad.
If such a facile solution did not occur to him, perhaps there must be something unsound in our logic.
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0 is a rational number
rational no. + rational no. = rational no.
(rational no.)(irrational no.)=irrational no.
a is rational
b.
=(rational no.)(irrational no.) = irrational no. , it can be only rational no. iff b=0
c.
=(rational no.)(irrational no.) = irrational no. , it can be only rational no. iff c=0
a+ (0) + (0) = 0
hence a=0
so we get solution as a=b=c=0