Prove cosh 3x = 4 cosh^3 x - 3 cosh x.

Prove we are to prove  4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. 
i.e. e^(3x)/2 + e^(-3x)/2 = 2 
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0 
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3, 
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 
= e^(3x) /2 + e^(-3x) /2 
= cosh(3x) 
= LHS 

Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. 

i.e. e^(3x)/2 + e^(-3x)/2 = 2 

Setting e^(3x) = u, we have u^2 + 1 - 4u = 0 

u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, 
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3, 

Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)