maths
posted by JOSHI .
write expansions:
1) (4/3x+3/4y)^3

in general
(a+b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3
then ...
(4/3x+3/4y)^3
= (4/3)^3 x^3 + 3(4/3)^2 x^2 (3/4)y + 3(4/3)x (3/4)^2 y^2 + (3/4)^3 y^3
= 64x^3/27 + 144x^2 y/36 + 108xy^2/48 + 27y^3/64
= 64x^3/27 + 4x^2y + 9xy^2/4 + 27y^3/64
testing by letting x=1, y=1
(4/3 + 3/4)^3 = appr 9.0422..
my expression at the end = .0422..
I assume you meant it the way you typed it
if you meant
( 4/(3x) + 3/(4x) )^3 it would totally change the answer.