Please help with simplification?

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

how do you get rid of all the fraction exponents, what's the common factor?

Answer 1

Numerator , N
(x + 4)^(-1/4) (x + 3)^(- 2/3) [(x + 4) + (x + 3)]
[ (x + 4)^(-1/4) ] [(x + 3)^(- 2/3)] (2x + 7)

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

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

Answer 2

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

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

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

Answer 3

negative power means 1/ power..
eg.. (x+3)^(-2/3) = 1/(x+3)^(2/3)

power square = multiply the power
[(x+4)^(3/4)]^2 = (x+4)^(6/4)

base^power1 * base^power2 = base ^(power1 + power2)
x^3 * x^4 = x^(3+4) = x^9

[(x+3)^(-2/3) * (x+4)^(3/4) - (x+3)^(1/3) * (x+4)^(-1/4)] / [(x+4)^(3/4)]^2
= [(x+4)^(3/4) / (x+3)^(2/3) - (x+3)^(1/3) / (x+4)^(1/4)] / (x+4)^(6/4)
= [(x+4)^{(3/4)+ 1/4} - (x+3)^{(1/3) + 2/3}] / (x+3)^(2/3) (x+4)^(1/4) (x+4)^(6/4)
= [(x+4) - (x+3)] / (x+3)^(2/3) (x+4)^(1/4 + 6/4)
= 1 / (x+3)^(2/3) (x+4)^(7/4)
or
= (x+3)^(-2/3) (x+4)^(-7/4)