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# simplify this! watch

1. please make solutions in white --->like this<---
if you can do this it feels very satisfying because there's a very subtle solution
ant try not to cheat by looking at others!

(I hope I've typed it out correctly)
Attached Images

2. Like this you mean? (Or do you want it simplified further or solved?)
Attached Images

3. dvs: shouldn't that be (n^2 + n)^(1/3) rather than (n^2 + 2n)^(1/3) ?
4. Yes, it should. I mistyped it!
5. I believe it can be simplifies further if I have types it out correctly.

hint:
factorise (a^3-b^3)
6. Using x^3 - y^3 = (x - y)(x^2 + xy + y^2),

1
= (n + 1) - n
= cbrt(n + 1)^3 - cbrt(n)^3
= [cbrt(n + 1) - cbrt(n)] [cbrt(n + 1)^2 + cbrt(n + 1)cbrt(n) + cbrt(n)^2]
= [cbrt(n + 1) - cbrt(n)] [cbrt(n^2 + 2n + 1) + cbrt(n^2 + n) + cbrt(n^2)]

(sum from n = 1 to 100) 1 / [cbrt(n^2 + 2n + 1) + cbrt(n^2 + n) + cbrt(n^2)]
= (sum from n = 1 to 100) [cbrt(n + 1) - cbrt(n)]
= cbrt(101) - 1
7. yeah that's right. another method:

a^2 + ab + b^2 = (a^3 - b^3)/(a - b)
1/(a^2 + ab + b^2) = (a - b)/(a^3 - b^3)

so 1/[3rt{1} + 3rt{2} + 3rt{4}] = (3rt 2 - 3rt 1)
ect. onto 1/[3rt{10000} + 3rt{10100} + 3rt{10201}] = (3rt 101 - 3rt 100)

all others in between cancel to leave 3rt 101 - 1

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Updated: February 3, 2005
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