$ x+y = 1 $ and $ \frac{1}{x} + \frac{1}{y} = 1 $ Solve $ x^3 + y^3 $

Question

$x$, $y$ are complex numbers, $x$ and $y$ aren't $0$.

$$ x + y = 1 $$ $$ \frac{1}{x} + \frac{1}{y} = 1 $$ $$ x^3 + y^3 = ? $$

Thank You!

For the complex numbers symbol, put in dollar signs \mathbb{C}.. For the real numbers it's \mathbb{R} etc.@ Siminore I think you mean $(x + y)(x^2 - xy + y^2)$ — Matthew Cassell, Jul 17 '15 at 13:12
@Axel I tried to find x and x = $ 1 - \frac{1}{y} $ then $ y^2 = 1 $ and then I have two options $ y = i $ or $ y = -i $ then I haven't continued to work because I have to choose between: A) 1 B) −2 C) −1 D) 8 as solution to $ x^3 + y^3 $ — ITChristian, Jul 17 '15 at 13:13
Related: http://math.stackexchange.com/questions/1349283/if-xy-1-x2y2-2-find-x7y7/1349294#1349294 — Dietrich Burde, Jul 17 '15 at 13:21

score 9 · Accepted Answer · answered Jul 17 '15 at 13:18

Multiplying the second equation by $xy$ we have $$x+y = 1 \quad \text{and} \quad xy = 1.$$ Any symmetric polynomial in two variables, such as $x^3 + y^3$, can be expressed in terms of $x+y$ and $xy$. Indeed, $$x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = 1 \cdot (1 - 3) = -2.$$

score 3 · Answer 2 · answered Jul 17 '15 at 13:25

3

\begin{align} \frac{1}{x}+\frac{1}{y}&=1 \\ \frac{x+y}{xy}&=1 \\ xy&=1 \\ (x+y)^3&=1^3 \\ x^3+y^3+3xy(x+y)&=1 \\ x^3+y^3&=-2 \end{align}

answered Jul 17 '15 at 13:25

g.kov

13,581

score 2 · Answer 3 · answered Jul 17 '15 at 13:19

A hint:

From the given data it is easy to compute the value of $xy$ as well. This means that you know the values both elementary symmetric functions $\sigma_1:=x+y$ and $\sigma_2:=xy$ of the unknown numbers $x$, $y\in{\mathbb C}$. This will allow you to compute any symmetric function of $x$ and $y$, in particular $\pi_3:=x^3+y^3$, without bothering about the values of $x$ and $y$ themselves. Look at Simiore's comment for hints.

$ x+y = 1 $ and $ \frac{1}{x} + \frac{1}{y} = 1 $ Solve $ x^3 + y^3 $

3 Answers3