You are here:

Advertisement

yeah sorry that was my bad. The index is x^(5/4).

(Simplify equation)

[(y^ 1/2)/(x^ 3/4) - (x^ 5/4)/(y^ 3/2)]^4

= {(y-x^2)/ [(x^ 3/4)(y^ 3/2)] }^4 (Combine both terms to achieve a common denominator)

= (y-x^2)^4 * 1/[(x^ 3/4)(y^ 3/2)] }^4

= (y-x^2)^4 * 1/[(x^ 3)(y^ 6)] }^4

At this point, you can obtain the binomial series expansion of (y-x^2)^4 , and combine with the product of 1/[(x^ 3)(y^ 6)] }^4 to attain your answer.

Hope it helps. Peace.

Algebra

Answers by Expert:

I can answer questions concerning calculus, complex numbers, vectors, statistics , algebra and trigonometry for the O level, A level and 1st/2nd year college math/engineering student.

More than 7 years of experience helping out in various homework forums. Latest presence is over at
http://www.thestudentroom.co.uk/.
You can also visit my main maths website http://www.whitegroupmaths.com where I have
designed "question locker" vaults to store tons of fully worked math problems.
A second one is currently being built.
Peace.**Organizations**

IEEE(Institute of Electrical and Electronics Engineers )**Education/Credentials**

Former straight As A level student from HCJC (aka HCI); scored distinctions
in both C and Further Mathematics
B Eng (Hons) From The National University Of Singapore (NUS)
B Sc (Hons) From University of London External (Grad Route)