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)