This post deals with the solution of cubic equation. The algebraic method of solving cubic equations is supposed to be due to the Italian, del Ferro (1465-1526). But it is called Cardano's method because it became known to people after the Italian, Girolamo Cardano, published it in 1545 in his 'Ars Magna'.
Omar Khayyam gave a great deal of thought to the cubic equations. Before him, Greek mathematicians obtained solutions for third degree equations by considering geometric methods that involved the intersection of conics. Let us consider the equation (ax^{3} + bx^{2} + cx + d = 0 ; a≠0)Step 1:Express ax^{3} + bx^{2} + cx + d = 0 as x^{3} + px^{2} + qx + r = 0Step 2:Shift the middle point of the curve on the axis x=−p/3.x^{3} + 3(p/3)x^{2} + 3(p^{2}/9)x + p^{3}/27 − 3(p^{2}/9)x − p^{3}/27 +qx+ r = 0(x + p/3)^{3} − [3(p^{2}/9) − q]x − p^{3}/27 + r = 0(x + p/3)^{3} − [p^{2}/3 − q]x − p^{3}/27 + r = 0Step 3: Let x + p/3 = y then x = y − p/3.Substitute y and (y − p/3) for (x + p/3) and x in the above equation.y ^{3} +[q − p^{2}/3](y − p/3) + r − p^{3}/27 = 0y ^{3} +[q − p^{2}/3]y − [qp/3 − p^{3}/9 − r + p^{3}/27] = 0y ^{3} +[q − p^{2}/3]y + [r − qp/3 + 2p^{3}/27] = 0y ^{3} +[q − p^{2}/3]y + [2p^{3}/27 − qp/3 + r] = 0Step 4:Now the equation is of the form y^{3} + Ay + B = 0where A = q − p ^{2}/3 and B = 2p^{3}/27 − qp/3 + rLet y = (s − A/3s) then s ^{3} − A^{3}/(3s)^{3} − As + A^{2}/3s + As − A^{2}/3s + B = 0s ^{3} − A^{3}/(3s)^{3} + B = 0Multiplying throughout by s ^{3}, we gets ^{6} + Bs^{3} − A^{3}/27 = 0Let, s ^{3} = zz ^{2} + Bz − A^{3}/27 = 0z = [−B ±√(B ^{2} + 4A^{3}/27)]/2s _{1}^{3} = [−B +√(B^{2} + 4A^{3}/27)]/2As we know that any equation has three cube roots. Let its cube root be s _{1}. then the roots are s_{1}, ωs_{1} and ω^{2}s_{1}. Similarly, s _{2}^{3} = [−B −√(B^{2} + 4A^{3}/27)]/2Let its cube root be s _{2}. then the roots are s_{2}, ωs_{2} and ω^{2}s_{2}.1/s _{1}^{3} = −(3s_{2}/A)^{3}⇒ 1/s _{1}^{3} = 1/[−B +√(B^{2} + 4A^{3}/27)]/2 = {[−B − √(B ^{2} + 4A^{3}/27)]/2}/{[−B −√(B^{2} + 4A^{3}/27)][−B +√(B^{2} + 4A^{3}/27)]/4}= {[−B − √(B ^{2} + 4A^{3}/27)]/2}/{−4A^{3}/(27×4)}= {[−B − √(B ^{2} + 4A^{3}/27)]/2}/{A^{3}/27)}= −(3s _{2}/A)^{3}s _{1}s_{2} = −A/3If we consider s _{1} then −A/3s_{1} = s_{2}.The product of s _{1} and s_{2} is real so the possibilities of the roots for y (= s − A/3s) ares _{1} + s_{2}, (ωs_{1} + ω^{2}s_{2}) and (ω^{2}s_{1} + ωs_{2}).Step 5:Shift the graph to the original position for the roots.The three roots ares _{1} + s_{2} − p/3, (ωs_{1} + ω^{2}s_{2} − p/3) and (ω^{2}s_{1} + ωs_{2} − p/3). The solution of the equationx^{3} + px^{2} + qx + r = 0 iss_{1} + s_{2} − p/3, (ωs_{1} + ω^{2}s_{2} − p/3) and (ω^{2}s_{1} + ωs_{2} − p/3).Where ω is a cube root of 1 i.e (−1 + i√3)/2 Where s_{1} is a cube root of [−B +√(B^{2} + 4A^{3}/27)]/2 and s_{2} is a cube root of [−B − √(B^{2} + 4A^{3}/27)]/2.and where A = q − p^{2}/3 and B = 2p^{3}/27 − qp/3 + r |

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