Factors, system and so on... $x+y=1$; $x^{2}+y^{2}=221$; what is $(x³)+(y³)$ ?
I already solved this problem, but I wanna know if there is an easier way to solve it, because I'm studying for a very hard exam and I must learn how to choose the best tactic to solve math problems. 
Check out what I did:
$x^{3}+y^{3}$
$(x+y)[x^{2}-xy+y^{2}]$
$(x+y)[x^{2}+2xy+y^{2}-3xy]$
$(x+y)[(x+y)^{2}-3xy]$
$(x+y)^{3}-3xy(x+y)$
$1^{3}-3xy1$
$1-3xy$
Then:
$x=1-y$
$(1-y)^{2}+y^{2}=221$
$1-2y+y^{2}+y^{2}=221$
$2y^{2}-2y-220=0$
$y=\frac{-(-2)±\sqrt{(-2)^{2}-8(-220)}}{4}$
$y’=(-10)$; $y''=11$
$x+y=1$ $\therefore$ $x=11; y=(-10)$
Then we have: 
$1-3*11(-10)=331$
And that's the answer!
My question is: There is an alternative way to solve this problem?

The posted proof is essentially correct, but note that the first half is not really necessary since you calculate the actual values of $x,y$ in the second half.
Regarding x=11; y=(-10) you should make a note that the system is symmetric in $x,y\,$, so $x=-10, y=11$ is a solution as well, but it doesn't matter which pair you choose since the expression to calculate ($x^3+y^3$) is itself symmetric in $x,y$.

There is an alternative way to solve this problem?



*

*$(x+y)^2=x^2+y^2+2xy \implies xy = \frac{1}{2}\left((x+y)^2-(x^2+y^2)\right)=\frac{1}{2}(1-221)=-110$

*$(x+y)^3=x^3+y^3+3xy(x+y) \implies x^3+y^3 = (x+y)^3-3xy(x+y) = 1 + 3 \cdot 110$
