<s>[INST] 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?
 [/INST] 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?



*

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

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