Cubic equations

The constant term is 6, and the leading coefficient is 1. Possible rational roots are the divisors of 6: $\pm 1, \pm 2, \pm 3, \pm 6$.

Let $P(x) = x^3 - 2x^2 - 5x + 6$.
Test $x=1$:

Since $P(1)=0$, $x=1$ is a root, and $(x-1)$ is a factor.

Dividing $x^3 - 2x^2 - 5x + 6$ by $(x-1)$:

1 | 1  -2  -5   6
  |    1  -1  -6
  ----------------
    1  -1  -6   0

The resulting quadratic equation is $x^2 - x - 6 = 0$.

Factor the quadratic equation:

This gives the roots $x=3$ and $x=-2$.

The solutions for the equation $x^3 - 2x^2 - 5x + 6 = 0$ are: $\boxed{x=1, x=3, x=-2}$

2) Solve $x^3 - 6x^2 + 11x - 6 = 0$

The constant term is -6, and the leading coefficient is 1. Possible rational roots are the divisors of -6: $\pm 1, \pm 2, \pm 3, \pm 6$.

Let $P(x) = x^3 - 6x^2 + 11x - 6$.
Test $x=1$:

Since $P(1)=0$, $x=1$ is a root, and $(x-1)$ is a factor.

Dividing $x^3 - 6x^2 + 11x - 6$ by $(x-1)$:

1 | 1  -6   11  -6
  |    1  -5    6
  ----------------
    1  -5    6   0

The resulting quadratic equation is $x^2 - 5x + 6 = 0$.

Factor the quadratic equation:

This gives the roots $x=2$ and $x=3$.

The solutions for the equation $x^3 - 6x^2 + 11x - 6 = 0$ are: $\boxed{x=1, x=2, x=3}$