this one is bugging me would appreciate any insights:

x^ {1/3}y^{1/2}(2x^{4/3}y^2-4x^{2/3}y^{-1/2}

thank you kindly

Question

this one is bugging me would appreciate any insights:    

x^ {1/3}y^{1/2}(2x^{4/3}y^2-4x^{2/3}y^{-1/2}

thank you kindly

anonymous · Answer

$$x^ {1/3}y^{1/2}(2x^{4/3}y^2-4x^{2/3}y^{-1/2}$$

anonymous · Answer

this much powers

anonymous · Answer

sorry i don't know how to make it bigger

hba · Answer

haah lol :P

anonymous · Answer

$$x^ {1/3}y^{1/2}(2x^{4/3}y^2-4x^{2/3}y^{-1/2})$$

jiteshmeghwal9 · Answer

$$\LARGE{{x^ {1/3}y^{1/2}(2x^{4/3}y^2-4x^{2/3}y^{-1/2})}}$$

anonymous · Answer

i know lol.. i think I can do this but just trying to read it is making my eyes crossed. :\

hba · Answer

@miszery Try solving it.We are here if you go wrong.

jiteshmeghwal9 · Answer

yup

anonymous · Answer

$$2x ^{5/3}y ^{5/2}-4x$$

anonymous · Answer

i got  $$2x^{5/3}y^{5/2}-4x^{3/3}y^0$$

anonymous · Answer

then sorry i type slow with the equaitons

anonymous · Answer

$$y ^{0}=1$$

hba · Answer

Due to exponentional law,

$$\huge\ e^0=1$$

Also,

$$3/3=1$$

anonymous · Answer

yep so  $$2x^{2/3}y^{2 {1/2} }$$

anonymous · Answer

crap can't type it :(

hba · Answer

Type it in normal or use the draw tool.

anonymous · Answer

-2^2/3y^2 1/2

anonymous · Answer

check it guys its solved in shorted form

anonymous · Answer

aw my answer is different from khurram :( i'm even more confused hehe

anonymous · Answer

i did a mistake dude... neglect it nd go n ahead

anonymous · Answer

ah ok thank you :D

anonymous · Answer

can anynone tell me if my last answer is in the right direction please :)    and also how do i give medals?