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Since  x + y  =  3     --->     y  =  3 - x

Since  xy  =  3     --->     y  =  3/x

 

Combining these equations:  3/x  =  3 - x     --->     3  =  3x - x²     --->     x² - 3x + 3  =  0

 

Use the quadratic formula to get:  x  =  ( 3 + i·sqrt(3) ) / 2  and  x  =  ( 3 - i·sqrt(3) ) / 2

 

[If x is one of the answers, then y is the other.]

 

Now, cube the answers to get the final answer ...  (it's an interesting answer) ...

Notice that \((x+y)^3=x^3+y^3+3xy(x+y)\). Then we plug in 3 for the (x+y)'s and xy's.

 

\((x+y)^3=x^3+y^3=3xy(x+y)\\ 3^3=x^3+y^3=3(3)(3)\\ 27=x^3+y^3=27\\ x^3+y^3=0\)

I am here to provide a fun alternative solution.

 

x + y = xy = 3 means that x and y are the roots of the quadratic equation \(t^2 - 3t + 3 = 0\).

(You can get this result by using Vieta's formula.)

 

Now, substituting x and y into the equation, \(\begin{cases}x^2 - 3x + 3 = 0\\y^2 - 3y + 3=0\end{cases}\), which means \(\begin{cases}x^2 = 3(x - 1)\\y^2 = 3(y - 1)\end{cases}\).

 

Using the result above:

\(\quad x^3 + y^3 \\= x(\color{red}x^2\color{black}) + y(\color{blue}y^2\color{black}) \\= \color{red}3\color{black}x\color{red}(x - 1)\color{black} + \color{blue}3\color{black}y\color{blue}(y - 1)\color{black} \\= 3x^2 + 3y^2 - 3(x + y)\)

 

Substituting x^2 = 3(x - 1) and y^2 = 3(y - 1) again and using the fact that x + y = 3,

\(\quad x^3 + y^3\\ = 3\color{red}x^2\color{black} + 3\color{blue}y^2\color{black} - 3(x + y)\\ = 3(\color{red}3(x - 1)\color{black}) +3(\color{blue}3(y - 1)\color{black}) - 3(x+y)\\ = 9x - 9 + 9y - 9 - 3(x + y) \\= 9(x + y) - 18 - 3(x + y)\\ =6(\color{purple}x + y\color{black}) - 18\\ =6(\color{purple}3\color{black}) - 18\\ =0\)