Question 241072
{{{2+root(3,13x-11-3x^2) = x}}}
<pre>

Isolate the cube root:

{{{root(3,13x-11-3x^2) = x-2}}}

Raise both sides to the 3rd power (that is, "cube" both sides)

{{{(root(3,13x-11-3x^2))^3 = (x-2)^3}}}

Raising a radical to the same exponent as its index amounts
to removing the radical and the exponent, so we have:

{{{13x-11-3x^2 = (x-2)^3}}}

But we must multiply the right side out:

{{{13x-11-3x^2 = (x-2)(x-2)(x-2)}}}

{{{13x-11-3x^2 = (x^2-4x+4)(x-2)}}}

{{{13x-11-3x^2 = x^3-2x^2-4x^2+8x+4x-8}}}

{{{13x-11-3x^2 = x^3-6x^2+12x-8}}}

Get 0 on one side:

{{{0=x^3-3x^2-x+3}}}

We like the 0 to be on the right, so we switch sides:

{{{x^3-3x^2-x+3=0}}}

Factor {{{x^2}}} out of the first two terms on the left:

{{{x^2(x-3)-x+3=0}}}

Factor {{{-1}}} out of the last two terms on the left:

{{{x^2(x-3)-1(x-3)=0}}}

Factor {{{(x-3)}}} out of both terms on the left:

{{{(x-3)(x^2-1)=0}}}

Factor the second parentheses as the difference of sqwuares:

{{{(x-3)(x-1)(x+1)=0}}}

Use the zero-factor principle and set each factor = 0:

x-3=0 gives solution x=3

x-1=0 gives solution x=1

x+1=0 gives solution x=-1

The solutions are -1, 1, and 3.  They all check in the original.

(You only need to check radical equations for extraneous solutions
when there is a root with an even index or a denominator containg
a variable.)

Edwin</pre>