A Simplified Closed-Form Expression for the General Cubic Equation Complete Derivation, Analysis, and Validation

A Simplified Closed-Form Expression for the General Cubic Equation: Complete Derivation, Analysis, and Validation Ajith Kumar K K Email: [email protected] October 30, 2025 Abstract We present a simplified closed-form expression for solving the general cubic equation ax3+bx2+cx +d= 0 that addresses the symbolic indeterminacy inherent in Cardano’s formula. Our method introduces symmetric parameters R,S, Tand an auxiliary variable tthat transform the cubic into a form yielding direct real-valued computation. We provide complete rigorous mathematical derivation with all algebraic steps, comprehensive analysis of degenerate cases, and numerical validation against standard methods. Testing on 1000 random polynomials demonstrates agreement to machine precision while avoiding intermediate complex arithmetic for real roots. The formula offers both theoretical elegance through its symmetric structure and practical advantages in specific computational applications. 1 Problem Statement Consider the general cubic equation: ax3+bx2+cx +d= 0,(a6= 0). Our goal is to find a closed-form expression for the roots that: 1. Avoids unnecessary complex arithmetic when roots are real 2. Handles degenerate cases without special branching logic 3. Provides clear algebraic structure amenable to analysis 2 Complete Derivation of the Simplified Formula 2.1 Transformation by Substitution We begin with the substitution x=y+t, where tis an auxiliary parameter to be determined. 1 Proposition 1. The substitution x=y+ttransforms the general cubic equation into: ay3+Ry2+Sy +T= 0, where the parameters are defined as: R= 3at +b, (1) S= 3at2+ 2bt +c, (2) T=at3+bt2+ct +d. (3) Proof. Substituting x=y+tinto the original equation: a(y+t)3+b(y+t)2+c(y+t) + d= 0. Expanding the powers: (y+t)3=y3+ 3y2t+ 3yt2+t3, (y+t)2=y2+ 2yt +t2, (y+t) = y+t. Substituting these expansions: a(y3+ 3y2t+ 3yt2+t3) + b(y2+ 2yt +t2) + c(y+t) + d= 0. Distributing: ay3+ 3aty2+ 3at2y+at3+by2+ 2bty +bt2+cy +ct +d= 0. Grouping by powers of y: ay3+ (3at +b)y2+ (3at2+ 2bt +c)y+ (at3+bt2+ct +d)=0. This yields the stated form with R,S, and Tas defined. 2.2 Reciprocal Transformation Lemma 1. The reciprocal transformation y=1 zconverts the equation ay3+Ry2+Sy + T= 0 into: Tz3+Sz2+Rz +a= 0. Proof. Substituting y=1 zinto the transformed cubic: a1 z3 +R1 z2 +S1 z+T= 0. This becomes: a z3+R z2+S z+T= 0. Multiplying through by z3: a+Rz +Sz2+T z3= 0. Rearranging: Tz3+Sz2+Rz +a= 0. 2 2.3 Determining the Auxiliary Variable t The key insight is to choose tsuch that a specific symmetry condition holds. Theorem 1 (Symmetry Constraint).To ensure elimination of the linear term in subsequent transformations, we require: R=S2 3T.(4) This constraint leads to a quadratic equation in t. Proof. From the constraint R=S2 3T, we have: 3at +b=(3at2+ 2bt +c)2 3(at3+bt2+ct +d). Multiply both sides by 3T= 3(at3+bt2+ct +d): 3(3at +b)(at3+bt2+ct +d) = (3at2+ 2bt +c)2. Expanding the left side: LHS = 3(3at +b)(at3+bt2+ct +d) = 3[3a2t4+ 3abt3+ 3act2+ 3adt +abt3+b2t2+bct +bd] = 3[3a2t4+ 4abt3+ (3ac +b2)t2+ (3ad +bc)t+bd] = 9a2t4+ 12abt3+ (9ac + 3b2)t2+ (9ad + 3bc)t+ 3bd. Expanding the right side: RHS = (3at2+ 2bt +c)2 = 9a2t4+ 12abt3+ 4b2t2+ 6act2+ 4bct +c2 = 9a2t4+ 12abt3+ (6ac + 4b2)t2+ 4bct +c2. Equating coefficients: Setting LHS = RHS and canceling 9a2t4+ 12abt3: (9ac + 3b2)t2+ (9ad + 3bc)t+ 3bd = (6ac + 4b2)t2+ 4bct +c2. Collecting terms: [(9ac + 3b2)−(6ac + 4b2)]t2+ [(9ad + 3bc)−4bc]t+ [3bd −c2] = 0 (3ac −b2)t2+ (9ad −bc)t+ (3bd −c2) = 0. This is our quadratic equation for t. 2.4 Solution for the Auxiliary Parameter Theorem 2. The auxiliary parameter tis given by: t=bc −9ad ±√∆ 2(3ac −b2),(5) where the discriminant is: ∆ = (9ad −bc)2−4(3ac −b2)(3bd −c2).(6) 3 Proof. Applying the quadratic formula to (3ac −b2)t2+ (9ad −bc)t+ (3bd −c2) = 0: Let A= 3ac −b2,B= 9ad −bc,C= 3bd −c2. Then: t=−B±√B2−4AC 2A=−(9ad −bc)±p(9ad −bc)2−4(3ac −b2)(3bd −c2) 2(3ac −b2). Expanding the discriminant: ∆ = (9ad −bc)2−4(3ac −b2)(3bd −c2) = 81a2d2−18abcd +b2c2−4(9abcd −3ac3−3b3d+b2c2) = 81a2d2−18abcd +b2c2−36abcd + 12ac3+ 12b3d−4b2c2 = 81a2d2−54abcd −3b2c2+ 12ac3+ 12b3d. Therefore: ∆ = 81a2d2−54abcd −3b2c2+ 12ac3+ 12b3d . (7) 2.5 Further Transformation Starting from Tz3+Sz2+Rz +a= 0, we make another substitution. Proposition 2. The substitution z=p+qtransforms the equation into: Tp3+ (3Tq +S)p2+ (3Tq2+ 2Sq +R)p+ (Tq3+Sq2+Rq +a) = 0. Proof. Substituting z=p+q: T(p+q)3+S(p+q)2+R(p+q) + a= 0. Expanding: (p+q)3=p3+ 3p2q+ 3pq2+q3, (p+q)2=p2+ 2pq +q2, (p+q) = p+q. Therefore: T(p3+ 3p2q+ 3pq2+q3) + S(p2+ 2pq +q2) + R(p+q) + a= 0. Expanding and grouping by powers of p: Tp3+ (3Tq +S)p2+ (3Tq2+ 2Sq +R)p+ (Tq3+Sq2+Rq +a) = 0. 4 2.6 Eliminating the Linear Term Theorem 3. To eliminate the coefficient of pin the transformed equation, we require: 3Tq2+ 2Sq +R= 0. Under the constraint R=S2 3T, this yields: q=−S 3T.(8) Proof. Solving 3Tq2+ 2Sq +R= 0 using the quadratic formula: q=−2S±√4S2−12TR 6T=−S±√S2−3TR 3T. Substituting the constraint R=S2 3T: q=−S±qS2−3T·S2 3T 3T=−S±√S2−S2 3T=−S±0 3T. Therefore: q=−S 3T. 2.7 Solving for p Theorem 4. With q=−S 3T, the equation for preduces to: Tp3+ (Tq3+Sq2+Rq +a)=0, yielding: p=3 r−2S3+ 9RST −27aT2 27T3.(9) Proof. Substituting q=−S 3Tinto the coefficients: Coefficient of p2: 3Tq +S= 3T·−S 3T+S=−S+S= 0. Coefficient of p: 3Tq2+ 2Sq +R= 0 (by design). Constant term: Tq3+Sq2+Rq +a=T−S 3T3 +S−S 3T2 +R−S 3T+a =T·−S3 27T3+S·S2 9T2−RS 3T+a =−S3 27T2+S3 9T2−RS 3T+a. 5 Finding common denominator 27T2: =−S3+ 3S3−9RST + 27aT2 27T2 =2S3−9RST + 27aT2 27T2. The equation becomes: Tp3+2S3−9RST + 27aT2 27T2= 0. Solving for p3: p3=−2S3−9RST + 27aT2 27T3. Therefore: p=3 r−2S3+ 9RST −27aT2 27T3. 2.8 Deriving the Final Expression Theorem 5 (Main Result).The root of the cubic equation ax3+bx2+cx +d= 0 is given by: x=3T −S+3 √−2S3+ 9RST −27aT2+t, (10) where tis given by equation (??), and R,S,Tare defined by equations (??), (??), (??). Proof. Recall the chain of substitutions: x=y+t, y=1 z, z=p+q. Working backwards: z=p+q=3 r−2S3+ 9RST −27aT2 27T3+−S 3T. Simplifying: z= 3 √−2S3+ 9RST −27aT2 3T−S 3T= 3 √−2S3+ 9RST −27aT2−S 3T. Then: y=1 z=3T 3 √−2S3+ 9RST −27aT2−S. Finally: x=y+t=3T 3 √−2S3+ 9RST −27aT2−S+t. This can be rewritten as: x=3T −S+3 √−2S3+ 9RST −27aT2+t . 6 3 Analysis of Degenerate Cases 3.1 Case 1: b2−3ac = 0 When b2−3ac = 0, the denominator in equation (??) vanishes, resulting in division by zero. However, the quadratic equation for tdegenerates to a linear equation that can be solved directly. Theorem 6. When b2= 3ac, the quadratic for tdegenerates to a linear equation with solution: t=c2−3bd 9ad −bc, and this leads to R= 0, S= 0. The final solution becomes: x=−1 b 3 pc(3bd −c2)−c b. Proof. When b2−3ac = 0, the coefficient of t2in the quadratic equation vanishes: (3ac −b2)t2+ (9ad −bc)t+ (3bd −c2) = 0 becomes: 0·t2+ (9ad −bc)t+ (3bd −c2) = 0. This is a linear equation in t: (9ad −bc)t=−(3bd −c2) = c2−3bd. Therefore: t=c2−3bd 9ad −bc. Substitute a=b2 3c: t=c2−3bd 9b2 3cd−bc =c2−3bd 9b2d 3c−bc =c2−3bd 3b2d−bc2 c . Simplify by multiplying numerator and denominator by c: t=c(c2−3bd) 3b2d−bc2=c(c2−3bd) b(3bd −c2). Since (c2−3bd) = −(3bd −c2), we get: t=c[−(3bd −c2)] b(3bd −c2)=−c b. Therefore, t=−c b. Verification that R= 0: 7 Substituting t=−c binto the expression for R: R= 3at +b = 3a·−c b+b =−3ac b+b =−3ac +b2 b. Since we have the condition b2= 3ac: R=−3ac + 3ac b=0 b= 0. Verification that S= 0: Substituting t=−c binto the expression for S: S= 3at2+ 2bt +c = 3a·−c b2 + 2b·−c b+c = 3a·c2 b2−2c+c =3ac2 b2−c. Using the condition b2= 3ac: S=3ac2 3ac −c=c−c= 0. Computing T: Substituting t=−c binto the expression for T: T=at3+bt2+ct +d =a−c b3 +b−c b2 +c−c b+d =a·−c3 b3+b·c2 b2−c2 b+d =−ac3 b3+c2 b−c2 b+d =−ac3 b3+d. Using b2= 3ac, we have b3=b·b2=b·3ac = 3abc: T=−ac3 3abc +d=−c2 3b+d=−c2+ 3bd 3b. Deriving the final expression for x: With R= 0 and S= 0, equation (??) becomes: x=3T −0 + 3 √−2·03+ 9 ·0·0·T−27aT2+t. 8 Simplifying: x=3T 3 √−27aT2+t. Note that: 3 √−27aT2=3 √−27a·3 √T2=−33 √a·T2/3. Therefore: x=3T −33 √a·T2/3+t=T1/3 −3 √a+t=−3 rT a+t. Substituting T=−c2+3bd 3band t=−c b: x=−3 r1 a·−c2+ 3bd 3b−c b. Using a=b2 3cfrom b2= 3ac: x=−3 r3c b2·3bd −c2 3b−c b. Simplifying: x=−3 rc(3bd −c2) b3−c b=−1 b 3 pc(3bd −c2)−c b. Therefore, the final solution for Case 1 is: x=−1 b 3 pc(3bd −c2)−c b. 3.2 Case 2: b=c= 0 When both b= 0 and c= 0, the original cubic equation simplifies significantly. Theorem 7. When b=c= 0, the cubic reduces to ax3+d= 0 with solution: x=−3 rd a. Our formula correctly reduces to this result. Proof. With b=c= 0, the original equation becomes: ax3+d= 0. This has the immediate solution: x3=−d a,thus x=−3 rd a. Verification using our formula: 9