how do u simplify the following expression?

Question

(x+4)^(3/4) times (x+3)^(-2/3) minus (x+3)^(1/3) times (x+4)^(-1/4) all divided by [(x+4)^(3/4)]^2

please help as i am not sure where to even begin with this one

Answer 1

Let N =
(x + 4)^(3/4) (x + 3)^(- 2/3) -
(x + 3)^(1/3)(x + 4)^(-1/4)

N = (x + 4)^(-1/4) (x + 3)^(-2/3) [ (x + 4) - (x + 3) ]
N = (x + 4)^(-1/4) (x + 3)^(-2/3) ( 1 )

D = (x + 4)^(3/2)

N / D = (x + 4)^(-7/4) (x + 3)^(-2/3)
N / D = 1 / [ (x + 4)^(7/4) (x + 3)^(2/3) ]

Answer 2

Writing the numerator as fractions
= (x+4)^3/4/(x+3)^2/3 - (x+3)^1/3/(x+4)1/4 then get common denominators
=[(x+4)^3/4(x+4)^1/4 - (x+3)^1/3(x+3)2/3]/(x+3)2/3(x+4)^1/4
=x+4-(x+3)/(x+3)2/3(x+4)^1/4 = 1/(x+3)2/3(x+4)^1/4
the denominator [(x+4)^3/4]² = (x+4)^6/4 or written as 1/ it can be multiplied
=1/(x+3)2/3(x+4)^1/4 times 1/(x+4)^6/4
=1/(x+3)2/3(x+4)^7/4