Wednesday, July 20, 2016

Solve for real solutions for the system: $x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$

Solve for real solutions for the system:

$x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$

My solution:

Let $x,\,y$ and $z$ be the real roots for a cubic polynomial.

From the relation $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, we have:

$(-a)^2=a^2+2(xy+yz+zx)$

$a^2=a^2+2(xy+yz+zx)\implies xy+yz+zx=0$

From the relation $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$, we have:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$

$-a^3-3xyz=(-a)(a^2-0)$

$-a^3-3xyz=-a^3\implies xyz=0$

We can now form the cubic polynomial in $t$ where its roots are $x,\,y$ and $z$:

$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz=0$

$t^3-(-a)t^2+(0)t-0=0$

$t^2(t+a)=0$

Obviously $t=0$ is the repeated root and the other root is $t=-a$.

Therefore we get the solution:

$(x,\,y,\,z)=(0,\,0,\,-a),\,(0,\,-a,\,0),\,(-a,\,0,\,0)$

2 comments:

  1. thoroughly steps to the solution. Nice job!

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  2. Thanks Michelle! It has always been my goal to let those even without the prerequisite knowledge to understand my blog posts, and that can only be achieved with the solid and good laid out of the solution. :D

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