Draw the angle bisector of A.

Draw a line segment $AD$ from vertex $A$ to side $BC$ such that $AD$ bisects $\angle A$. This means $\angle BAD \cong \angle CAD$.

Now we have two triangles: $\triangle ABD$ and $\triangle ACD$.

In $\triangle ABD$ and $\triangle ACD$:

• $\angle B \cong \angle C$ (Given)
• $\angle BAD \cong \angle CAD$ (By construction, $AD$ is the angle bisector of $\angle A$)
• $AD \cong AD$ (Common side to both triangles)

Therefore, by the AAS congruence criterion, $\triangle ABD \cong \triangle ACD$.

Since the two triangles are congruent, their corresponding parts are congruent (CPCTC).
Thus, the side $AB$ in $\triangle ABD$ corresponds to the side $AC$ in $\triangle ACD$.
Hence, $AB \cong AC$.

This proves that if two angles of a triangle are congruent, then the sides opposite to them are also congruent.

We look for two numbers that multiply to 24 and add up to -11. These numbers are -3 and -8.

The polynomial $x^2 - 6x + 9$ is a perfect square trinomial, which can be factored as $(x-3)^2$.

The product of two polynomials is equal to the product of their HCF and LCM:

Where $L(x)$ is the LCM. We can rearrange this to find $L(x)$:

Cancel one $(x-3)$ term from the numerator and the denominator.

We can expand this expression:

The LCM of the two polynomials is $\boxed{(x-8)(x-3)^2}$ or $\boxed{x^3 - 14x^2 + 57x - 72}$.