Question 460861
{{{a^3-b^3=(a-b)(a^2+ab+b^2)}}}

{{{a^2+ab+b^2=((a+(b/2))^2+3b^2/4)}}}

but if {{{a>b}}} then {{{a-b>0}}} and clearly {{{a+(b/2)^2>0}}} and {{{3b^2/4>0}}}

so, {{{(a-b)(a^2+ab+b^2)>0}}}, hence {{{a^3-b^3>0}}} and {{{a^3>b^3}}}