# 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 $$x$$?

• Here's a question. What happens when you expand out $(x-2.5)^2$? (This is not necessary for solving the problem.) Jan 30, 2023 at 1:25

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:

$$\large{\displaystyle{\color{purple}{x}^3 + \color{purple}{y}^3 =(\color{red}{x + y})^3 - 3\color{green}{xy}\big(\color{red}{x+y}\big)}}$$

\begin{aligned} x^3+y^3 & =(x+y)^3-3 x y(x+y) \\ & =5^3-3(1)(5) \\ & =110 \end{aligned}

• On the money, neat. Jan 30, 2023 at 8:11

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: $$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)((x+y)^2-3xy)$$

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.

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 $$x$$.