Question

How many terms will be there in the expansion of
\[\large (\sqrt{x}+1)^{8}-(\sqrt{x}-1)^{8}\]
??

Answer 1

\[\large (\sqrt{x}+1)^{8}-(\sqrt{x}-1)^{8}\]

Answer 2

bionomial expansion is screaming out to me here.
things are going to cancel out too.

Answer 3

youre going to get 9 terms in the first expansion, and another 9 terms in the second expansion
now, for the second expansion, the negatives are going to be attached to every odd term
these in turn will cancel out with the corresponding equal terms of the first expansion, hence leaving the even terms
furthermore , you'll have 4 terms xD
this is probably wrong xD

Answer 4

yep i was right, its 4 terms. i am a genius
http://www.wolframalpha.com/input/?i=expand++%28sqrt%7Bx%7D%2B1%29%5E%7B8%7D-%28sqrt%7Bx%7D-1%29%5E%7B8%7D%29

Answer 5

I'll do a few,
For \(\large ( \sqrt{x}+1)^{8}\),
-------------------------------------
\[^8C_0(1)^0(\sqrt{x})^8=x^4\]
\[^8C_1(1)^1(\sqrt{x})^7=8x^{7/2}\]
\[^8C_2(1)^2(\sqrt{x})^6=28x^3\]
\[^8C_3(1)^3(\sqrt{x})^5=56x^{5/2}\]
-------------------------------------
So, \(\large ( \sqrt{x}-1)^{8}\), you can see that the sign alternates,

Instead of getting
\[x^4+8x^{7/2}+28x^3+56x^{5/2}\]
You will get
\[x^4-8x^{7/2}+28x^3-56x^{5/2}\]
and so on.
So you don't need to expand the whole thing, but instead you expand one of the brackets only. Then for sure the other bracket's sign alternates.

Answer 6

both has 9 terms ...
the left has 5 + ve terms and 4 -ve terms ...
these - (5 +ve ) should cancel the 5 terms of preceding expansion

so you have 8 terms ... which can be added into compact 4 terms!!

Answer 7

we have (sqrt(x) + 1)^8 - (sqrt(x) -1)^8
(sqrtx + 1)^2 = (sqrtx -1)^2 + 4sqrtx
on that substitution ,
we have
( (sqrtx -1)^2 + 4sqrtx )^4 - (sqrtx -1)^8
expansion of (a+b)^4 has 5 terms..
here,,1 term gets cancelled out on expansion which is (sqrtx -1)^8
so you are left with 4 terms..