Is this correct please - Binomial Theorem

Natasha1
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I'm a complete beginner at these:

(3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)^2 + 1

which would give me

= 3 (9n^3 + 6n^2) +1 is this correct?
 
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anyone :confused: ??
 
No, that is not correct.
(3n)^3 + 3(3n)^2 + 3(3n)^2 + 1
Your exponents should go from 3 down to 0.
3 (9n^3 + 6n^2) +1
This is also not correct.
 
Natasha1 said:
I'm a complete beginner at these:

(3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)^2 + 1
Perhaps it is a typo but you should not have "^2" twice.

which would give me

= 3 (9n^3 + 6n^2) +1 is this correct?
You have added 3(3n)^2 and 3(2n)^2 but the second shouldn't be squared.
 
thanks :-)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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