Finding $x^3+y^3$, given $x + y = 5$ and $xy = 1$ 
Given
$$x + y = 5 \qquad xy = 1$$
Find $x^3 + y^3$.

To solve this, I tried this:
$y = \frac{1} {x}$
$x + \frac{1}{x} = 5$
$x^2 - 5x + 1 = 0$
What formula needs to be used to find the value of $x$?
 A: Another method that requires some intuition and knowledge of cubics:
Observe that, $$\large{\displaystyle{(\color{red}{x + y})^3 = \color{purple}{x}^3 + 3x^2y + 3xy^2 + \color{purple}{y}^3} = \color{purple}{x}^3 + 3\color{green}{xy}\big(\color{red}{x+y}\big) + \color{purple}{y}^3}$$
Otherwise re-arranged, gives:
$$\large{\displaystyle{\color{purple}{x}^3 + \color{purple}{y}^3 =(\color{red}{x + y})^3 - 3\color{green}{xy}\big(\color{red}{x+y}\big)}}$$
A: $$
\begin{aligned}
x^3+y^3 & =(x+y)^3-3 x y(x+y) \\
& =5^3-3(1)(5) \\
& =110
\end{aligned}
$$
A: You can use the quadratic formula, or use a trick. Recall that $x^3+y^3=(x+y)(x^2-xy+y^2) $.
Now consider that $x^2-xy+y^2=(x+y)^2-3xy$
So you have:
$$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)((x+y)^2-3xy)  $$
A: Note that $x^3+y^3=(x+y)(x^2-xy+y^3)$. We have $x+y=5$. Then $(x+y)^2=x^2+y^2+2xy=25$. Then, $x^2-xy+y^2=25-3xy=22$. All that is left is to multiply them.
A: Use the binomial formula. We have
$$(x+y)^3=x^3 + 3x^2y + 3xy^2 + y^3,$$
which reduces to
$$125=x^3 + 3\cdot 5 + y^3$$
and thus
$$x^3 + y^3 = 110.$$
There is no need to solve for $x$.
