Solving nonlinear systems without using systems

x+y=1

xy=1

x^{3}+y^{3}=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 >.<

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 $\displaystyle \displaystyle \begin{align*} x = -1^{\frac{1}{3}} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} y = -1 - \left(-1^{\frac{1}{3}}\right) \end{align*}$, then

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

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

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

Checking the discriminant reveals that there are not any real solutions.

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.

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?

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

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

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

$\displaystyle \displaystyle \begin{align*} x^3 + y^3 &= 1\left[ 1^2 - 3(1) \right] \\ &= -2 \end{align*}$

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.

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.

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

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

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

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

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

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