1. **State the problem:** Prove the identity $$2\cosh(3x)\cosh(x) = \cosh(4x) + \cosh(2x)$$. 2. **Recall the formula for hyperbolic cosine product:** $$\cosh(a)\cosh(b) = \frac{\cosh(a+b) + \cosh(a-b)}{2}$$ 3. **Apply the formula to the left side:** $$2\cosh(3x)\cosh(x) = 2 \times \frac{\cosh(3x+x) + \cosh(3x - x)}{2} = \cosh(4x) + \cosh(2x)$$ 4. **Conclusion:** The left side simplifies exactly to the right side, so the identity is proven. This shows that $$2\cosh(3x)\cosh(x) = \cosh(4x) + \cosh(2x)$$ is true for all real $x$.