Expert: Paul Klarreich - 10/1/2011

Question
I was looking at the question and answer you gave on http://en.allexperts.com/q/Advanced-Math-1363/2008/10/Solution-cube-roots.htm#b
Can you please explain how you go from first step to next here..
z^3 = 4 + sqrt(15) + 4 - sqrt(15) + 3(4 + sqrt(15))^2/3(4 - sqrt(15))^1/3 + 3(4 + sqrt(15))^1/3(4 - sqrt(15))^2/3

z^3 = 8 + 3(4 + sqrt(15))^1/3(16 - 15)^1/3 + 3(16 - 15)^1/3(4 - sqrt(15))^1/3
When I calculate it out I end up making all in brackets total 1, there for z^3=14 ..
however I am unsure of how you re-arranged it..

If we take the first part which you called a..
3(4 + sqrt(15))^2/3(4 - sqrt(15))^1/3
I see we can simplify it by calculation to get
3(16 - 4(sqrt(15)) + 4(sqrt(15)) - 15

Please can you show me how to get it to
3(4 + sqrt(15))^1/3(16 - 15)^1/3
I know multiplying numbers that have power of 1/3 and 2/3 totals

many thanks
Elli

Answer
Questioner:Elli
Country:London, City of, United Kingdom
Category:Advanced Math
Private:No
Subject: law of indices

Question:

I was looking at the question and answer you gave on http://en.allexperts.com/q/Advanced-Math-1363/2008/10/Solution-cube-roots.htm#b
Can you please explain how you go from first step to next here..
z^3 = 4 + sqrt(15) + 4 - sqrt(15) + 3(4 + sqrt(15))^2/3(4 - sqrt(15))^1/3 + 3(4 + sqrt(15))^1/3(4 - sqrt(15))^2/3

z^3 = 8 + 3(4 + sqrt(15))^1/3(16 - 15)^1/3 + 3(16 - 15)^1/3(4 - sqrt(15))^1/3
When I calculate it out I end up making all in brackets total 1, there for z^3=14 ..
however I am unsure of how you re-arranged it..

If we take the first part which you called a..
3(4 + sqrt(15))^2/3(4 - sqrt(15))^1/3
I see we can simplify it by calculation to get
3(16 - 4(sqrt(15)) + 4(sqrt(15)) - 15

Please can you show me how to get it to
3(4 + sqrt(15))^1/3(16 - 15)^1/3
I know multiplying numbers that have power of 1/3 and 2/3 totals

many thanks
Elli
-----------------------------------------
(I will be writing  s for sqrt(15) to save typing.)

If z  = (4 + s)^1/3 + (4 - s)^1/3

Then in the binomial expansion:
(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3;

We can  use  a = (4 + s)^1/3  and  b = (4 - s)^1/3

Now  a^3 = ((4 + s)^1/3)^3 = 4 + s
and  b^3 = ((4 - s)^1/3)^3 = 4 - s,

so a^3 + b^3 = 8.

That is two of the terms. The other two are:

3a^2 b = 3((4 + s)^1/3)^2 (4 - s)^1/3
3a b^2 = 3(4 + s)^1/3 ((4 - s)^1/3)^2

3a^2 b = 3(4 + s)^1/3 (4 + s)^1/3 (4 - s)^1/3
3a b^2 = 3(4 + s)^1/3 (4 - s)^1/3 (4 - s)^1/3

3a^2 b = 3[ (4 + s) (4 + s) (4 - s) ]^1/3
3a b^2 = 3[ (4 + s) (4 - s) (4 - s) ]^1/3


Now, since (4 + s)(4 - s) = (16 - s^2),
we can write:

3a^2 b = 3[ (4 + s) (16 - s^2) ]^1/3
3a b^2 = 3[ (16 - s^2) (4 - s) ]^1/3

And, since  s^2 = 15,  [remember what 's' stands for]

16 - s^2 = 1 (!!!!)

3a^2 b = 3(4 + s)^1/3
3a b^2 = 3(4 - s)^1/3

and the sum is:

3a^2 b + 3a b^2 = 3(4 + s)^1/3 + 3(4 - s)^1/3
3a^2 b + 3a b^2 = 3[ (4 + s)^1/3 + 3(4 - s)^1/3 ]

Then the entire expansion becomes:

z^3 = 8 + 3[(4 + s)^1/3 + (4 - s)^1/3]

and the thing inside the [] is exactly z.

Does that do it for you?