(a^3+b^3)^2=(a^2+b^2)^3 find (a/b)+(b/a) please give the solution

dude just expand both the sides..... v get on simplification 2ab = 3(a^2)+3(b^2)....... then take 2ab on the rhs then cancelling common terms..... v get 1=3/2(a/b + b/a) i.e., 2/3= a/b + b/a.... u got???? is the answer correct??? gimme a reply....

That`s right.thanks

Dude I think you are right :)

(a^3+b^3)^2=(a^2+b^2)^3
then (a^3+b^3)^2=a^6[1+(b/a)^2]^3
(a^3+b^3)^2 / (a^6)=[1+(b/a)^2]^3
(1+(b/a)^3)^2=[1+(b/a)^2]^3
1+2(b/a)^3+(b/a)^6=1+3(b/a0^2+3(b/a)^4+(b/a)^6
cancelling 1+(b/a)^6 both sides we get
3(b/a)^2-2(b/a)^3+3(b/a)^4=0
dividing by (b/a)^3
3(b/a+a/b)=2
then a/b+b/a=2/3
thanks and regards
M.MURALIKRISHNA
askIITIANS FACULTY

a^6 + b^6 + 2a^3. b^3 = a^6 + b^6 + 3a^4 b^2 + 3 a^2 b^4

Thanks and Regards,
M.MURALIKRISHNA
askIITians faculty