Cubic Equation by Hyperbolic method

While search for the solution of Quintic equation I found methods to find solution of polynomial equations with the help of hyperbolic functions. The complex roots occur in pairs ( R (cos θ + i sin θ) and R(cos θ - i sin θ)) and and can be represented as exponents (Reθ and Re). The hyperbolic cosine function (cosh k = (ek + e-k)/2) is function of sum of two exponential functions. So, they can be used to solve polynomial equations. The complex roots occur in pairs as conjugates and a hyperbolic cosine function is sum of two reciprocal exponential functions.

For example:
cosh 3k = cosh 2k cosh k + sinh 2k sinh k
= (cosh2 k + sinh2 k) cosh k + 2 sinh k cosh k sinh k
= (2cosh2 k - 1) cosh k + 2 sinh2 k cosh k
= (2cosh3 k - cosh k + 2 (cosh2 k - 1) cosh k
= 4 cosh3 k - 3cosh k

(cosh3 k)= (1/4)(3 cosh k + cosh 3k)

Express x3 + b'x2 + c'x + d' = 0 as x3 = -cx -d

As we know that in this form the sum of three roots is zero so we can express one root as negative of the sum of other two roots. So we can assume the roots to be Rek and Re-k.
Then the solution is -R(ek + e-k).
and -2R cosh k = -R(ek + e-k)

Making the equation
(cosh3 k)= (1/4)(3 cosh k + cosh 3k)

equivalent to the cubic equation by multiplying throughout by -8R3, we get
(-8R3 cosh3 k)= (1/4)(-24 R3 cosh k - 8R3 cosh 3k)
(-2R cosh k)3= 3R2 (-2R cosh k) - 2R3cosh 3k

We get,
3R2 = -c
and 2R3 cosh 3k = d
R = (-c/3)1/2
cosh 3k = d/[2(-c/3)3/2]
[e3k + e-3k]/2 = d/[2(-c/3)3/2]
[e6k + 1] = {d/[(-c/3)3/2]}e3k
e6k -{d/(-c/3)3/2}e3k + 1 = 0
Solving we get, e3k = {[d/(-c/3)3/2] ±√{[d/(-c/3)3/2]2 - 4}}/2
e3k = [1/(-c/3)3/2]{d ±√[d2 - 4(-c/3)3]}/2
e3k = [1/(-c/3)3/2]{d/2 ±√[(d/2)2 - (-c/3)3]}
ek = {[1/(-c/3)3/2]{d/2 +√[(d/2)2 - (-c/3)3]}}1/3
e-k = {[1/(-c/3)3/2]{d/2 -√[(d/2)2 - (-c/3)3]}}1/3
Rek = {d/2 +√[(d/2)2 - (-c/3)3]}1/3
Re-k = {d/2 -√[(d/2)2 - (-c/3)3]}1/3

solution is
-R(ek + e-k) = -{d/2 +√[(d/2)2 - (-c/3)3]}1/3 - {d/2 -√[(d/2)2 - (-c/3)3]}1/3
 = {(-d/2) -√[(d/2)2 - (-c/3)3]}1/3 + {(-d/2) +√[(d/2)2 - (-c/3)3]}1/3
= {(-d/2) -√[(d/2)2 + (c/3)3]}1/3 + {(-d/2) +√[(d/2)2 + (c/3)3]}1/3
 Hence solution is {(-d/2) -√[(d/2)2 + (c/3)3]}1/3 + {(-d/2) +√[(d/2)2 + (c/3)3]}1/3



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