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I disagree with explanations given above. Here is my approach:

Stmt (1) indicates that x=y otherwise x^6-y^6=0 does not hold true. Hence, x^3-y^3 is always equal to zero. SUFF.

Stmt (2) is clearly INSUFF because there is no inf. about x.

If my approach is wrong, I would appreciate any other explanations.

Yes, your approach is wrong: x^6-y^6=0 implies that either x=y or x=-y, for example 1^6-1^6=0 and also 1^6-(-1)^6=0. To see this algebraically you could rewrite x^6-y^6=0 as (x^3-y^3)(x^3+y^3)=0 --> either x^3=y^3, or x^3=-y^3 --> so either x=y or x=-y. OR x^6-y^6=0 --> x^6=y^6 --> x^2=y^2 --> x=y or x=-y.

Now, if x=y then x^3-y^3=0 for any values of x and y BUT of x=-y then x^3-y^3=2x^3 and we need the value of x (or y) the get the single numerical value of 2x^3. So statement (1) is not sufficient.

(2) y=0 --> clearly insufficient.

(1)+(2) y=0, so x=0 too (as x^6-y^6=0) and x^3-y^3=0. Sufficient.

This is not tough question, but you need to be carefull not to fall into trap of ODD POWER of numbers. you always need to be suspicious when you see one. _________________

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