1. ## binomial problem

HI,
I need help with this one (x^1/2 - y^1/2)^5.

just need to know where to start.

Thanks.

2. ## Re: binomial problem

Originally Posted by anthonye
HI,
I need help with this one (x^1/2 - y^1/2)^5.

just need to know where to start.

Thanks.
the formula for a binomial expansion is probably a good place to start

$(x+y)^n=\sum_{k=0}^n \left(n \atop k\right)x^k y^{n-k}\mbox{ where }\left(n \atop k\right)=\frac{n!}{k!(n-k)!}$

note that $(x-y)=(x+(-y))$

3. ## Re: binomial problem

ok but what do I do with the powers on the variables?

4. ## Re: binomial problem

Originally Posted by anthonye
ok but what do I do with the powers on the variables?
$(x^{1/2} - y^{1/2})^n = \left((x^{1/2}) + (-y^{1/2})\right)^n$

5. ## Re: binomial problem

Just not seeing a method here?

6. ## Re: binomial problem

Go to the formula. Wherever you see an $x$, replace it by $(x^{1/2})$. Wherever you see a $y$, replace it by $(-y^{1/2})$.

Example: $x^k$ becomes $(x^{1/2})^k$

7. ## Re: binomial problem

OK but wouldn't the first term be 5c0(x^1/2)^5.

8. ## Re: binomial problem

Yes, that is correct. $(x^{1/2})^5 = x^{1/2}\cdot x^{1/2} \cdot x^{1/2} \cdot x^{1/2} \cdot x^{1/2} = x^{1/2+1/2+1/2+1/2+1/2} = x^{5\cdot 1/2} = x^{5/2}$. In other words, $(a^b)^c = a^{b\cdot c}$.