Question 935200
if {{{x +y = a}}} and {{{xy = b}}}, then 

{{{x = a-y}}}and {{{y = b/x}}}

the value of {{{1/x^3 + 1/y^3}}}

= {{{1/(a-y)^3 + 1/(b/x)^3}}}

={{{1/(a-y)^3 + x^3/b^3}}}

={{{(b^3 + x^3(a-y)^3)/((a-y)^3b^3)}}}

={{{((ax+b-xy)(a^2x^2-abx-2ax^2y+b^2+bxy+x^2y^2))/(b^3(a-y)^3)}}}

={{{(a^3x^3-3a^2x^3y+3ax^3y^2+b^3-x^3y^3)/(a^3b^3-3a^2b^3y+3ab^3y^2-b^3y^3)}}}