# Tedious Simplification: Show expression is real

1. Mar 29, 2012

### JolleJ

After doing A LOT of simplification on a complicated expression I am now at a point where my own skills can't take me any further.

My problem is that I have an expression, which I am almost certain must be real, but it contains the imaginary units here and there. I have tried using Maple and Mathematica with different assumptions, but it does not simplify very much.

Now, I do not expect you to do my tedious algebra. However, I was hoping that one of you might know of a technique (in Maple/Mathematica/Hand) that could help me?

My expression is this:

${\frac {\sqrt [3]{2}n \left( -1+n \right) \left( -1+2\,n \right) \left( 2\,\sqrt [3]{-2}+4\,\sqrt [3]{-2} \left( -1+n \right) n+ \left( -1-2\,n \left( 1+ \left( -3+n \right) n \right) +i\sqrt {1-16 \,n+44\,{n}^{2}-44\,{n}^{3}+28\,{n}^{4}-24\,{n}^{5}+12\,{n}^{6}} \right) ^{2/3}-i\sqrt {3} \left( -1-2\,n \left( 1+ \left( -3+n \right) n \right) +i\sqrt {1-16\,n+44\,{n}^{2}-44\,{n}^{3}+28\,{n}^{4 }-24\,{n}^{5}+12\,{n}^{6}} \right) ^{2/3} \right) }{\sqrt [3]{-1-2\,n \left( 1+ \left( -3+n \right) n \right) +i\sqrt {1-16\,n+44\,{n}^{2}- 44\,{n}^{3}+28\,{n}^{4}-24\,{n}^{5}+12\,{n}^{6}}} \left( -8+16\,n \right) }}$
where n is an interger larger than 1.

as you can see the same terms appear many places.

Any help will be greatly appreciated.

Maple format:
Code (Text):
2^(1/3)*n*(-1+n)*(-1+2*n)*(2*(-2)^(1/3)+4*(-2)^(1/3)*(-1+n)*n+(-1-2*n*(1+(-3+n)*n)+I*sqrt(1-16*n+44*n^2-44*n^3+28*n^4-24*n^5+12*n^6))^(2/3)-I*sqrt(3)*(-1-2*n*(1+(-3+n)*n)+I*sqrt(1-16*n+44*n^2-44*n^3+28*n^4-24*n^5+12*n^6))^(2/3))/((-1-2*n*(1+(-3+n)*n)+I*sqrt(1-16*n+44*n^2-44*n^3+28*n^4-24*n^5+12*n^6))^(1/3)*(-8+16*n))
Mathematica format:
Code (Text):
(2^(1/3) (-1 + n) n (-1 + 2 n) (2 (-2)^(1/3) +
4 (-2)^(1/3) (-1 + n) n + (-1 - 2 n (1 + (-3 + n) n) +
I Sqrt[1 - 16 n + 44 n^2 - 44 n^3 + 28 n^4 - 24 n^5 +
12 n^6])^(2/3) -
I Sqrt[3] (-1 - 2 n (1 + (-3 + n) n) +
I Sqrt[1 - 16 n + 44 n^2 - 44 n^3 + 28 n^4 - 24 n^5 +
12 n^6])^(2/3)))/((-8 + 16 n) (-1 - 2 n (1 + (-3 + n) n) +
I Sqrt[1 - 16 n + 44 n^2 - 44 n^3 + 28 n^4 - 24 n^5 + 12 n^6])^(
1/3))

Last edited: Mar 29, 2012
2. Mar 29, 2012

### Robert1986

Here's one idea:

If the number is pure real, then you can multiply the i's by -1 and after this, you should have the same number as before.

3. Mar 29, 2012

### JolleJ

Thanks. Yes, that is a good idea. However, the expression is so complicated that it doesn't really help. I can't show the expressions are the same.

And actually, which I should have clarified, I do need the simplified expression.