Question 1208446
<pre>

{{{matrix(2,1,"",(x^2 + 4)^(4/3) + x * (4/3)(x^2 + 4)^(1/3) * 2x)}}}

Factor out {{{matrix(2,1,"",(x^2 + 4)^(1/3))}}}

{{{matrix(2,1,"",(x^2 + 4)^(1/3)((x^2 + 4)^(3/3) + expr(4/3)2x^2))}}} 

{{{matrix(2,1,"",(x^2 + 4)^(1/3)((x^2 + 4)^1 + expr(8/3)*x^2))}}}

{{{matrix(2,1,"",(x^2 + 4)^(1/3)(x^2 + 4 + expr(8/3)*x^2))}}}

Get the fraction on the outside of the parentheses by writing everything
over the least common denominator of 3, then factor out 1/3"

{{{matrix(2,1,"",(x^2 + 4)^(1/3)(expr(3/3)*x^2 + 12/3 + expr(8/3)*x^2))}}} 
 
{{{matrix(2,1,"",expr(1/3)(x^2 + 4)^(1/3)*(3x^2 + 12 + 8x^2))}}} 

{{{matrix(2,1,"",expr(1/3)(x^2 + 4)^(1/3)*(11x^2 + 12))}}} 

Ikleyn will probably say it's not simplified if I don't change the
1/3 power to a cube root, so here goes:

{{{matrix(2,1,"",expr(1/3)*root(3,x^2+4)(11x^2 + 12))}}}

Now all she can complain about is that I don't skip steps and write 3/3
and then 1 exponents.

Edwin</pre>