Question

prove cosh 3x = 4 cosh^3 x - 3 cosh x by definition

Answer 1

Using the definition, you're trying to prove:
\[\frac{ e^{3x}+e^{-3x} }{ 2 }=4(\frac{ e^{x}+e^{-x} }{ 2 })^{3} - 3(\frac{ e^{x}+e^{-x} }{ 2 })\]So the easiest thing to do is to manipulate the right-hand side and try to get it to match the left. As a start, I would rewrite as:
\[\frac{ e^{3x}+e^{-3x} }{ 2 }= [\frac{ e^{x}+e^{-x} }{ 2 }][4(\frac{ e^{x}+e^{-x} }{ 2 })^{2}-3]\]
So try foiling out those terms on the right side and you should be able to get it to match the left if done properly :)

Answer 2

you could also use a^3 + b^3 formula :

\[\frac{ e^{3x}+e^{-3x} }{ 2 }=\dfrac{(e^x)^3 +(e^{-x})^3 }{2} = \dfrac{(e^x+e^{-x})^3 - 3 (e^x+e^{-x})}{2} = \cdots\]

Answer 3

Dat be fancy, hehe :D

Answer 4

:) a^3 + b^3 = (a+b)^3 - 3ab(a+b)