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We are given the equation:

\[
x^2 - m \equiv (x - 3)(x + n) - 3
\]

### Step 1: Expand the right-hand side
Let's first expand the right-hand side expression \((x - 3)(x + n)\):

\[
(x - 3)(x + n) = x^2 + nx - 3x - 3n = x^2 + (n - 3)x - 3n
\]

Now, subtract 3 from the expanded expression:

\[
(x^2 + (n - 3)x - 3n) - 3 = x^2 + (n - 3)x - 3n - 3
\]

Thus, the equation becomes:

\[
x^2 - m \equiv x^2 + (n - 3)x - 3n - 3
\]

### Step 2: Compare both sides
We now have:

\[
x^2 - m \equiv x^2 + (n - 3)x - 3n - 3
\]

For the equation to hold true for all \(x\), the terms involving \(x\) and the constant terms must be equal.

- The \(x^2\) terms cancel out on both sides.
- There is no linear term in \(x\) on the left-hand side, so the coefficient of \(x\) on the right must be zero:
  
  \[
  n - 3 = 0 \implies n = 3
  \]

- The constant terms on both sides must also be equal. On the left-hand side, the constant is \(-m\), and on the right-hand side, the constant is \(-3n - 3\). Substituting \(n = 3\) into this:

  \[
  -m = -3(3) - 3 = -9 - 3 = -12
  \]

Thus, \(m = 12\).

### Final Answer:
The value of \(m\) is \(12\).

---

Do you need further details or explanations? Here are five questions to deepen your understanding:

1. How does expanding a binomial product help solve equations like this?
2. Why is it important to compare the coefficients of similar terms when solving an equation?
3. Can this method be used for more complex equations? Why or why not?
4. How would the solution change if different constants were involved?
5. What would happen if we had a quadratic term on the right-hand side?

**Tip:** When comparing polynomials, always make sure to match coefficients of corresponding powers of \(x\). This simplifies the process of finding unknown constants.

If m and n are constants such that x^2 - m ≡ (x - 3)(x + n) - 3, find the value of m.

Learn how to solve for the constant m in the equation x^2 - m ≡ (x - 3)(x + n) - 3 with a step-by-step solution. This problem involves key algebraic concepts such as expanding binomials and comparing coefficients.

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