Solve the given expression for the sum of the two numbers:
[tex] x+y = \dfrac{x^3+y^3}{x^2-xy+y^2} [/tex]
which you can rewrite as
[tex] x+y = \dfrac{x^3+y^3}{x^2+y^2-xy} [/tex]
Now, we are given the product of the numbers (i.e. xy) to be 10, the sum of the squares of the numbers (i.e. [tex] x^2+y^2 [/tex]) to be 29, and the sum of the cubes of the numbers (i.e. [tex] x^3+y^3 [/tex]) to be 133.
If we plug these values in the formula written above, we have
[tex] x+y = \dfrac{133}{29-10} = \dfrac{133}{19} = 7 [/tex]
