Solve:
$\displaystyle x^3-y^3=602$
$\displaystyle xy(y-x)=-198$
Answer:
x = 11 and y = 9
or
x = -9 and y = -11
$\displaystyle x^3-y^3=602$
$\displaystyle -3x^2y+3xy^2=-594$
If we combine the two expressions, we get a perfect cube:
$\displaystyle x^3 - 3x^2y+3xy^2-y^3=602-594 \implies (x-y)^3=8 \implies x - y =\ldots$
This is your key to solving for x and y. Can you do it now?