# Solve simultaneous eqs. `x^3=5x+y` ; `y^3=x+5y` ?

embizze | Certified Educator

Solve the system `x^3=5x+y,y^3=x+5y` :

Solving the first equation for `y` we get `y=x^3-5x`. Substituting into the second equation we get `(x^3-5x)^3=x+5(x^3-5x)` .

Expanding and collecting like terms we get:

`x^9-15x^7+75x^5-125x^3=x+5x^3-25x`

`x^9-15x^7+75x^5-130x^3+24x=0`

`x(x^8-15x^6+75x^4-130x^2+24)=0`

Thus x=0,y=0 is one solution. We solve the 8th degree polynomial:

Using the rational root theorem we find -2 and 2 are roots. Using synthetic division or polynomial long division we have:

`x^8-15x^6+75x^4-130x^2+24`

`=(x-2)(x+2)(x^6-11x^4+31x^2-6)=0`

There are no rational roots of `x^6-11x^4+31x^2-6` , so we try `+-sqrt(6)` which work. Using synthetic division we get :

`x^6-11x^4+31x^2-6=(x^2-6)(x^4-5x^2+1)`

We treat `x^4-5x^2+1` as a quadratic in `x^2` ; using completing the square we get:

`x^4-5x^2+1=0 ==> (x^2-5/2)^2-25/4+1=0`

`(x^2-5/2)^2=21/4`

`x^2-5/2=+-sqrt(21)/2`

`x^2=(5+-sqrt(21))/2`

`x=+-sqrt(1/2(5+-sqrt(21)))`

Thus the x values for the solution are `-2,0,2,-sqrt(6),sqrt(6),+-sqrt(1/2(5+-sqrt(21)))`

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The solutions to the system are:

`(0,0),(-2,2),(2,-2),(-sqrt(6),-sqrt(6)),(sqrt(6),sqrt(6))`

`(-sqrt(1/2(5-sqrt(21))),1/2sqrt(1/2(5-sqrt(21)))(5+sqrt(21)))`

`(sqrt(1/2(5-sqrt(21))),-1/2sqrt(1/2(5-sqrt(21)))(5+sqrt(21))`

`(-sqrt(1/2(5+sqrt(21))),-1/2sqrt(1/2(5-sqrt(21)))(sqrt(21)-5))`

`(sqrt(1/2(5+sqrt(21))),1/2sqrt(1/2(5-sqrt(21)))(sqrt(21)-5))`

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