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 (Re
^{θ} and Re^{-θ}). The hyperbolic cosine function (cosh k = (e^{k} + 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 = (cosh ^{2} k + sinh^{2} k) cosh k + 2 sinh k cosh k sinh k= (2cosh ^{2} k - 1) cosh k + 2 sinh^{2} k cosh k= (2cosh ^{3} k - cosh k + 2 (cosh^{2} k - 1) cosh k= 4 cosh ^{3} k - 3cosh k(cosh ^{3} k)= (1/4)(3 cosh k + cosh 3k)Express x ^{3} + b'x^{2} + c'x + d' = 0 as x^{3} = -cx -dAs 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 Re ^{k} and Re^{-k}.Then the solution is -R(e ^{k} + e^{-k}). and -2R cosh k = -R(e ^{k} + e^{-k})Making the equation (cosh ^{3} k)= (1/4)(3 cosh k + cosh 3k)equivalent to the cubic equation by multiplying throughout by -8R ^{3}, we get(-8R ^{3} cosh^{3} k)= (1/4)(-24 R^{3} cosh k - 8R^{3} cosh 3k)(-2R cosh k) ^{3}= 3R^{2} (-2R cosh k) - 2R^{3}cosh 3kWe get, 3R ^{2} = -cand 2R ^{3} cosh 3k = dR = (-c/3) ^{1/2}cosh 3k = d/[2(-c/3) ^{3/2}][e ^{3k} + e^{-3k}]/2 = d/[2(-c/3)^{3/2}][e ^{6k} + 1] = {d/[(-c/3)^{3/2}]}e^{3k}e ^{6k} -{d/(-c/3)^{3/2}}e^{3k} + 1 = 0Solving we get, e ^{3k} = {[d/(-c/3)^{3/2}] ±√{[d/(-c/3)^{3/2}]^{2} - 4}}/2e ^{3k} = [1/(-c/3)^{3/2}]{d ±√[d^{2} - 4(-c/3)^{3}]}/2e ^{3k} = [1/(-c/3)^{3/2}]{d/2 ±√[(d/2)^{2} - (-c/3)^{3}]}e ^{k} = {[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}Re ^{k} = {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(e ^{k} + 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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