+983 Hi Guest!
We use: \(x^3-y^3=(x-y)(x^2+xy+y^2) \)
We already know \( (x-y)\) and \(x^2+y^2\)
We can plug the known values in, and we get:
\(6(24+xy)=x^3-y^3\)
To solve for xy, we use: \((x-y)^2=x^2-2xy+y^2\)
This means that \(6^2=24-2xy\)
Solving for xy, we get:
\(36=24-2xy\\ 12=-2xy\\ xy=-6\)
Now we can plug this into the original expression:
\(6(24-6)=x^3-y^3\\ x^3-y^3=108\)
I hope this helped,
Gavin
