If there are cube roots of 6 on RHS, then they must appear on LHS somewhere, say the natural substitution:
((a)^(1/3)+ (6a)^(1/3)1)^2 =49+20*6^(1/3)
WolframAlpha gives a=48, from which Charlie's solutions can be derived.
Incidentally, the cube root of 48 is 2*6^(1/3) and of 288 is 2*6^(2/3)

Posted by broll
on 20150610 07:22:32 