# Solving nonlinear systems without using systems

• Aug 28th 2012, 06:25 PM
Greymalkin
Solving nonlinear systems without using systems
x+y=1
xy=1
x3+y3=z
wolfram gives me x=-1^(1/3), y=1-(-1^(1/3)), and z=-2, and another set.
Not only can't I solve this problem without crutching with wolfram, there is supposed to be an easier method to solve without using systems, what is this method?

PS sorry for no latex, I'm limited to cell phone Internet until I finish moving >.<
• Aug 28th 2012, 07:21 PM
Prove It
Re: Solving nonlinear systems without using systems
What do you mean you can't solve this problem without using "systems". This IS a system of equations.

Also, the answer you have given is incorrect. If \displaystyle \begin{align*} x = -1^{\frac{1}{3}} \end{align*} and \displaystyle \begin{align*} y = -1 - \left(-1^{\frac{1}{3}}\right) \end{align*}, then

\displaystyle \begin{align*} xy &= -1^{\frac{1}{3}}\left[ -1 - \left(-1^{\frac{1}{3}}\right) \right] \\ &= -1^{\frac{1}{3}}\left( -1 + 1^{\frac{1}{3}} \right) \\ &= 1^{\frac{1}{3}} - 1^{\frac{2}{3}} \\ &\neq 1 \end{align*}

which is required by the second equation.

This equation actually does not have any real solutions. From the first equation, we have y = 1 - x.

Substituting into the second equation gives

\displaystyle \begin{align*} x(1 - x) &= 1 \\ x - x^2 &= 1 \\ 0 &= x^2 - x + 1 \end{align*}

Checking the discriminant reveals that there are not any real solutions.
• Aug 28th 2012, 09:15 PM
Greymalkin
Re: Solving nonlinear systems without using systems
I meant the problem in my book stated that there was a way to solve without using the methods used to solve systems of equations, it is also supposed to be (sic)"easier", I am asked to find this secret method.
• Aug 28th 2012, 09:19 PM
Greymalkin
Re: Solving nonlinear systems without using systems
The problem by verbatim is:
The sum of two numbers is 1, and their product is 1. Find the sum of their cubes. There is a method to solve this
Problem that is easier than solving a nonlinear system of equations. Can you discover it?
• Aug 28th 2012, 09:34 PM
Prove It
Re: Solving nonlinear systems without using systems
There do not exist any two real numbers that have a sum of 1 and a product of 1, I showed you this already. Your question is nonsense.

Edit: If you are allowed to go into the complex numbers, since you are trying to find the sum of their cubes, note that

\displaystyle \begin{align*} x^3 + y^3 &= (x + y)\left( x^2 - x\,y + y^2 \right) \\ &= (x + y)\left( x^2 + 2\,x\,y + y^2 - 3\,x\,y \right) \\ &= (x + y)\left[ (x + y)^2 - 3\,x\,y \right] \end{align*}

And since you know \displaystyle \begin{align*} x + y = 1 \end{align*} and \displaystyle \begin{align*} x\,y = 1 \end{align*}, that gives

\displaystyle \begin{align*} x^3 + y^3 &= 1\left[ 1^2 - 3(1) \right] \\ &= -2 \end{align*}
• Aug 28th 2012, 10:09 PM
Greymalkin
Re: Solving nonlinear systems without using systems
I should have mentioned that the wolfram solutions were listed under complex. Thank you very much for your time.
• Aug 28th 2012, 10:27 PM
Soroban
Re: Solving nonlinear systems without using systems
Hello, Greymalkin!

Quote:

The sum of two numbers is 1, and their product is 1.
Find the sum of their cubes.

$\text{Let }x\text{ and }y\text{ be the two numbers.}$

$\text{We are told: }\:\begin{Bmatrix}x+y\:=\:1 & [1] \\ xy \:=\:1 & [2] \end{Bmatrix}$

$\text{Cube [1]: }\:(x+y)^3 \:=\:1^3 \quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:1$

$\text{We have: }\:(x^3+y^3) + 3xy(x+y) \:=\:1$

$\text{But }\,xy \,=\,1\,\text{ and }\,x+y\,=\,1$

. . $\text{So we have: }\:x^3 + y^3 + 3\!\cdot\!1\!\cdot\!1 \:=\:1 \quad\Rightarrow\quad \boxed{x^3+y^3 \:=\:-2}$