Show that cosh3x is equivalent to 4 cosh3x minus three coshx

Let $A = 2x$ and $B = x$.

$$\text{LHS} = \cosh(3x) = \cosh(2x + x) = \cosh(2x) \cosh x + \sinh(2x) \sinh x$$

Recall that $\cosh(2x) = 2 \cosh^2 x - 1$ and $\sinh(2x) = 2 \sinh x \cosh x$.

$$\text{LHS} = (2 \cosh^2 x - 1) \cosh x + (2 \sinh x \cosh x) \sinh x$$

Substitute this into the expression.

$$\text{LHS} = 2 \cosh^3 x - \cosh x + 2 (\cosh^2 x - 1) \cosh x$$

$$\text{LHS} = 2 \cosh^3 x - \cosh x + 2 \cosh^3 x - 2 \cosh x$$ $$\text{LHS} = (2 \cosh^3 x + 2 \cosh^3 x) + (-\cosh x - 2 \cosh x)$$ $$\text{LHS} = 4 \cosh^3 x - 3 \cosh x$$

Since LHS = RHS, the identity is proven.

The final answer is $\text{Shown}$.