The problem asks to find one root of the cubic equation $x^3 + 2x - 3x^2 - 6 = 0$ from the given options: $1, -1, -3, 3$.

1. Problem Description

The problem asks to find one root of the cubic equation $x^3 + 2x - 3x^2 - 6 = 0$ from the given options: $1, -1, -3, 3$.

2. Solution Steps

We can rewrite the equation as:

$x^3 - 3x^2 + 2x - 6 = 0$

Now, we test each of the given options to see which one satisfies the equation.

Option 1: $x = 1$

$(1)^3 - 3(1)^2 + 2(1) - 6 = 1 - 3 + 2 - 6 = -6 \neq 0$

So, $x = 1$ is not a root.

Option 2: $x = -1$

$(-1)^3 - 3(-1)^2 + 2(-1) - 6 = -1 - 3 - 2 - 6 = -12 \neq 0$

So, $x = -1$ is not a root.

Option 3: $x = -3$

$(-3)^3 - 3(-3)^2 + 2(-3) - 6 = -27 - 3(9) - 6 - 6 = -27 - 27 - 6 - 6 = -66 \neq 0$

So, $x = -3$ is not a root.

Option 4: $x = 3$

$(3)^3 - 3(3)^2 + 2(3) - 6 = 27 - 3(9) + 6 - 6 = 27 - 27 + 6 - 6 = 0$

So, $x = 3$ is a root.

Alternatively, we can try to factor the polynomial.

$x^3 - 3x^2 + 2x - 6 = 0$

$x^2(x - 3) + 2(x - 3) = 0$

$(x^2 + 2)(x - 3) = 0$

The roots are $x = 3$ and $x^2 = -2$ which gives $x = \pm i\sqrt{2}$.

The only real root from the options is $x = 3$.

3. Final Answer

3