2011/090) Prove that there do not exist four positive real numbers, all distinct from each other, such that a^3 + b ^3 = c^3 + d^3 and a+b = c+d

proof:
without loss of generality assume a>=b and c >= d

now as a+b = c+ d
cube both sides a^3+b^3 + 3ab(a+b)= c^3+d^3 + 3cd(c+d)

and a a^3+b^3 = c^3+d ^3 so 3ab(a+b) =3cd(c+d)
and hence ab = cd
(a-b)^2 = (a+b)^2 - 4ab = (c+d)^2 - 4cd = (c-d)^2 and hence a- b = c- d

as
a + b = c+ d
a- b = c - d
adding 2a = 2c = a = c and hence b = d
hence proved