\((x-y)=1\) square both sides
\(x^2-2xy+y^2=1\) (1)
given \(x^2+y^2=7\) substitute into (1)
\(7-2xy=1\)
Solve for \(xy\)
\(-2xy=-6\)
\(xy=3\)
\(x^3-y^3=(x-y)(x^2+xy+y^2)\) (The difference of two cubes)
Substitute all givens and the value of \(xy\)
\((1)(7+3)\)
\(1*10=10\)
Thus \(x^3-y^3=10\)
