Q13/112) Show that if a > b then a^3 > b^3

We have (a^3-b^3) = (a-b)(a^2+ab+b^2)

We need to show that (a^2+ab+b^2) > 0

We have a^2 + ab + b ^2 = (a-b)^2 + 3ab … (1)

a^2 + ab + b ^2 = (a+b)^2 – ab … (2)

as a > b if a or b is zero from (1) or (2) a^2 + ab + b^2 > 0

if ab > 0 then from (1) a^2 + ab + b ^2 > 0

and if ab < 0 then from (2) a^2 + ab + b ^2 > 0

 from above we have the result