If $x>y=x^3+y^3$ and $x@y=x^3-y^3$, then $3>(2@1)=$

Let's evaluate the custom operators step-by-step according to the definitions given:
1. Definition of $x>y=x^3+y^3$
2. Definition of $x@y=x^3-y^3$

First, let's calculate the inner bracket expression: $(2@1)$
Using the definition of $@$ with $x=2$ and $y=1$:

$$2@1=2^3-1^3=8-1=7$$

Now substitute this value back into the main expression:

$$3>(2@1)=3>7$$

Using the definition of $>$ with $x=3$ and $y=7$:

$$3>7=3^3+7^3=27+343=370$$

Thus, the final value is $370$.