The image contains the equation $x^3 - 6x^2 + 11x - 6 = 0$.

To solve this cubic equation, we can follow these steps:

1. Factorization by grouping or trial and error for possible rational roots: The Rational Root Theorem suggests that any rational solution to the polynomial equation $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0$ is a fraction $p/q$, where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$. For $x^3 - 6x^2 + 11x - 6 = 0$, the possible rational roots are the factors of $-6$ (constant term) divided by the factors of $1$ (leading coefficient). Therefore, the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$.

2. Testing possible roots:

   • Testing $x = 1$: $1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$ Since $x = 1$ is a root, we can factor $x - 1$ out of the polynomial.

3. Synthetic division: Divide $x^3 - 6x^2 + 11x - 6$ by $x - 1$ using synthetic division:

   1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array}$$ The quotient is $$x^2 - 5x + 6$$.

4. Solve the quadratic equation $x^2 - 5x + 6 = 0$: $x^2 - 5x + 6 = 0$ Factor the quadratic: $(x - 2)(x - 3) = 0$ Therefore, $x = 2$ and $x = 3$.

5. List all roots: The roots of the equation $x^3 - 6x^2 + 11x - 6 = 0$ are $x = 1, x = 2, x = 3$.

If you have any questions or need further details, please let me know. Here are five questions you might ask next:

1. How do you verify the roots of a polynomial equation?
2. Can you explain the Rational Root Theorem in more detail?
3. What other methods can be used to solve cubic equations?
4. How does synthetic division work for polynomials?
5. Can you show how to solve a cubic equation using the general formula?

Tip: When solving polynomial equations, always start by checking for possible rational roots using the Rational Root Theorem; it can significantly simplify the problem.