Originally Posted by **topsquark**

My basic attack is to get the equation that the expression represents. There may be a simpler way to do it, but this is the way I grok things...

$\displaystyle x=(6\sqrt{3}+10)^{1/3}-(6\sqrt{3}-10)^{1/3}$

Cubing both sides:

$\displaystyle x^3=(6\sqrt{3}+10)-3(6\sqrt{3}+10)^{2/3}(6\sqrt{3}-10)^{1/3}$

$\displaystyle +3(6\sqrt{3}+10)^{1/3}(6\sqrt{3}-10)^{2/3}-(6\sqrt{3}-10)$

Express the second term as:

$\displaystyle (6\sqrt{3}+10)^{1/3}(6\sqrt{3}+10)^{1/3}(6\sqrt{3}-10)^{1/3}$

Then multiply the last two factors in this term:

$\displaystyle (6\sqrt{3}+10)^{1/3}(108-100)^{1/3}=2(6\sqrt{3}+10)^{1/3}$

Doing this to the third term as well, and simplifying, we get:

$\displaystyle x^3=20-6(6\sqrt{3}+10)^{1/3}+6(6\sqrt{3}-10)^{1/3}$

Rearranging a bit we get:

$\displaystyle (x^3-20)/6=-(6\sqrt{3}+10)^{1/3}+(6\sqrt{3}-10)^{1/3}$

Now, at the beginning of this, we defined the LHS of this to be -x, so

$\displaystyle (x^3-20)/6=-x$

or,

$\displaystyle x^3+6x-20=0$

Graphing this, we can easily see there is only one real root. Then, using whatever method you choose, we can find that the real solution is x=2.

-Dan