simplifying (1/2)^(2/3)w_1^(1/3)w_2^(2/3)y + 2^(1/3)w_1^(1/3)w_2^(2/3)y = ...

[sadegs]

Hi,

sorry but I'm currently having a block and have no idea how to solve this. Can anyone show me how:

(1/2)^2/3w1^1/3w2^2/3y + 2^1/3w1^1/3w2^2/3y = 3[(w1w2²)/4]^1/3y

my main problem is how you get 3/4^1/3 from (1/2)^2/3+2^1/3


​Thanks in advance!

 

[Deleted member 4993]

Hi,

sorry but I'm currently having a block and have no idea how to solve this. Can anyone show me how:

(1/2)^2/3w1^1/3w2^2/3y + 2^1/3w1^1/3w2^2/3y = 3[(w1w2²)/4]^1/3y

my main problem is how you get 3/4^1/3 from (1/2)^2/3+2^1/3


​Thanks in advance!

(1/2)^2/3+2^1/3= 1/(2)^2/3 + 2^1/3 =[1 + 2^1/3*2^2/3]/(2^2/3) = ??

 

[JeffM]

One baby step at a time.

\(\displaystyle \left ( \dfrac{1}{2} \right )^{2/3} + 2^{1/3} =\)

\(\displaystyle \dfrac{1^{2/3}}{2^{2/3}} + 2^{1/3} =\)

\(\displaystyle \dfrac{1}{2^{2/3}} + 2^{1/3} =\)

\(\displaystyle \dfrac{1}{2^{2/3}} + 2^{1/3} * 1 = \)

\(\displaystyle \dfrac{1}{2^{2/3}} + 2^{1/3} * \dfrac{2^{2/3}}{2^{2/3}} = \)

\(\displaystyle \dfrac{1}{2^{2/3}} + \dfrac{2^{1/3} * 2^{2/3}}{2^{2/3}} =\)

\(\displaystyle \dfrac{1}{2^{2/3}} + \dfrac{2^{(1/3)+(2/3)}}{2^{2/3}} =\)

\(\displaystyle \dfrac{1}{2^{2/3}} + \dfrac{2^1}{2^{2/3}} =\)

\(\displaystyle \dfrac{1}{2^{2/3}} + \dfrac{2}{2^{2/3}} =\)

\(\displaystyle \dfrac{3}{2^{2/3}} =\)

\(\displaystyle \dfrac{3}{(2^2)^{1/3}} =\)

\(\displaystyle \dfrac{3}{4^{1/3}}.\)