Question

Expand the binomial:

Answer 1

\[(\sqrt{x}-\sqrt{y})^{5}\]

Answer 2

i will expand it

Answer 3

( x^1/2 - y^1/2 ) ^5 = we use pascal triangle coefficients 1 5 10 10 5 1

Answer 4

I am going to let a = x^1/2, and b = y^1/2 , then we can go back and substitute
( x^1/2 - y^1/2 ) ^5 = a^5 + 5a^4*b + 10a^3*b^2 + 10a^2*
b^3 + 5a*b^4 + b^5

Answer 5

\[(a-b)^{5} = a ^{5}-5a ^{4}b+10a ^{3}b ^{2}-10a ^{2}b ^{3}+5ab ^{4}-b ^{5}\]
now replace:
\[a = \sqrt{x} \]
\[b = \sqrt{y}\]

Answer 6

you did the same thing i did, wow. what a waste of time

Answer 7

and you used pretty print too.

Answer 8

@perl you forgot your negatives

Answer 9

that doesnt help me.....

Answer 10

@laurenbae what will help you?

Answer 11

5 i x^(3/2) sqrt(y) e^(i pi floor(-(arg(y))/(2 pi)))

see here
http://www.wolframalpha.com/input/?i=%28x%E2%88%9A%E2%88%92y%E2%88%9A%295&dataset=

Answer 12

but that's not one of the choices it gives me >.<

Answer 13

This might be an easier approach than the pascals triangle..
\[(a+b)^{n} = \binom{n}{k}*a^{n-k}*b^{k}\]
\(( \sqrt{x} - \sqrt{y})^5\)
\[ \sum_{k=0}^{5}\binom{5}{k} * \left(\sqrt{x}\right)^{5-k}
*\left(-\sqrt{y}\right)^{k}\]

Answer 14

Are you able to do it?

Answer 15

no which is why im asking for help. >.<

Answer 16

where are you having problems?