Question 1208447
<pre>

{{{matrix(2,1,"",6(6x + 1)^(1/3)(4x - 3)^(3/2) + 6(6x + 1)^(4/3)(4x - 3)^(1/2))}}}

let a = {{{matrix(2,1,"",(6x + 1)^(1/3))}}}
let b = {{{matrix(2,1,"",(4x - 3)^(1/2))}}}

{{{6ab^3+6a^4b}}}

{{{6ab(b^2+a^3)}}}

{{{matrix(2,1,"",6(6x + 1)^(1/3)(4x - 3)^(1/2)((4x - 3)^1+(6x + 1)^1)))}}}

{{{matrix(2,1,"",6(6x + 1)^(1/3)(4x - 3)^(1/2)(4x - 3+6x + 1)))}}}

{{{matrix(2,1,"",6(6x + 1)^(1/3)(4x - 3)^(1/2)(10x - 2)))}}}

{{{matrix(2,1,"",6(6x + 1)^(1/3)(4x - 3)^(1/2)(2(5x - 1))))}}}

{{{matrix(2,1,"",12(6x + 1)^(1/3)(4x - 3)^(1/2)(5x - 1)))}}}

You can leave it like that, or change the fraction powers to roots.

{{{matrix(2,1,"",12*root(3,6x + 1)*sqrt(4x - 3^"")(5x - 1)))}}}

You could even express the product of roots as a 6th root, but the
polynomial under the 6th root radical would make it too complicated.
So let's quit here.

Edwin</pre>