Question 1156961
Let x, y, z be nonzero real numbers such that {{{x + y + z = 0}}}, 
and {{{xy + xz + yz <> 0}}}. Find all possible values of 
{{{(x^5+y^5+z^5)/((xyz)(xy+xz+yz))}}}

<pre>
{{{x + y + z = 0}}}

{{{z=-x-y}}}

Substitute for z

{{{(x^5+y^5+(-x-y)^5)/((xy^""(-x-y))(xy^""+x(-x-y)+y(-x-y)))}}}

{{{((x+y^"")(x^4-x^3y+x^2y^2-xy^3+y^4)-(x+y)^5)/(-xy(x+y)(xy-x^2-xy-xy-y^2))}}}

{{{((x+y^"")(x^4-x^3y+x^2y^2-xy^3+y^4)-(x+y)^5)/(-xy(x+y)(-x^2-xy-y^2))}}}

{{{((x+y^"")(x^4-x^3y+x^2y^2-xy^3+y^4)-(x^""+y)^5)/(xy(x^""+y)(x^2+xy+y^2))}}}

Divide top and bottom by (x+y)

{{{((x^4-x^3y+x^2y^2-xy^3+y^4)-(x+y^"")^4)/(xy(x^2+xy+y^2))}}}

{{{((x^4-x^3y+x^2y^2-xy^3+y^4)-(x^4+4x^3y+6x^2y^2+4xy^3+y^4))/(xy(x^2+xy+y^2))}}}

{{{(x^4-x^3y+x^2y^2-xy^3+y^4-x^4-4x^3y-6x^2y^2-4xy^3-y^4)/(xy(x^2+xy+y^2))}}}

{{{(-5x^3y-5x^2y^2-5xy^3)/(xy(x^2+xy+y^2))}}}

{{{(-5xy(x^2+xy+y^2))/(xy(x^2+xy+y^2))}}}

{{{-5}}}

So -5 is the only value it can have.

Edwin</pre>