Pedroβs Method for Solving Cubic Equations. Pedro A. DR November 5, 2024 Introduction Consider the depressed cubic equation π₯ 3 + ππ₯ + π = 0 For real numbers, and . By the fundamental theorem of algebra, every π π cubic has three roots, which may be real or complex. The goal is to find the roots of our polynomial. Many methods have arisen since the first solution to the cubic found by Scipione del Ferro. Methods such as completing the cube, Cardanoβs method, Vietaβs substitution, Trigonometric solutions, Resolvents, and Tschirnhausβs work on transformations are valid ways to solve cubics. Introducing new parameters While these methods have proven useful, We will introduce a new method for solving cubic equations that builds on foundational concepts from previous methods but introduces additional complexity. The main idea is to set , π₯ = π’ + π£ where we plug into the depressed cubic and expand to get the following. π₯ π’ 3 + 3π’ 2 π£ + 3π’π£ 2 + π£ 3 + ππ’ + ππ£ + π = 0 Rearranging the cubic in a manner like so will give the following equation π’ 3 + 3π’ 2 π£ + (3π£ 2 + π)π’ + π£ 3 + ππ£ + π = 0 Here we multiply our arbitrary cubic by a new cubic factor π’ 3 + ππ’ 2 + ππ’ + π The idea behind this step is that we have introduced three new variables and the two cubic products result in a sextic. The three variables help us cancel out the terms, but we also have the extra variable, which can help us cancel π’ 5 , π’ 4 , π’ 2 π£ out the term which will leave us with a sextic where and π’ π’ 6 + Ξ±π’ 3 + Ξ² = 0 Ξ± Ξ² are constants built off of the depressed cubic. Multiplying the two cubic factors on the left hand side gives π’ 6 + (3π£ + π)π’ 5 + (3π£ 2 + 3π£π + π + π)π’ 4 + (π£ 3 + 3π£ 2 π + 3π£π + π + ππ + π + π£π)π’ 3 + (π£ 3 π + 3π£ 2 π + 3π£π + π£ππ + ππ + ππ) π’ 2 + (π£ 3 π + 3π£ 2 π + π£ππ + ππ + ππ)π’ + π( π£ 3 + ππ£ + π) = 0 Making four systems of equations and equating them to zero gives the following 3π£ + π = 0 3π£ 2 + 3π£π + π + π = 0 π£ 3 π + 3π£ 2 π + 3π£π + π£ππ + ππ + ππ = 0 π£ 3 π + 3π£ 2 π + π£ππ + ππ + ππ = 0 Solving for the first equation, we get that so plugging that into the three π =β 3π£ remaining equations gives β 6π£ 2 + π + π = 0 β 3π£ 4 + 3π£ 2 π + 3π£π β 3π£ 2 π + ππ β 3π£π = 0 π£ 3 π + 3π£ 2 π + π£ππ + ππ + ππ = 0 We can solve each equation for a variable, by solving the second equation for we π find out that , so plugging into the two remaining equations give π = 6π£ 2 β π 15π£ 4 + 3ππ£ β π 2 β 3π£π = 0 6π£ 5 + 5π£ 3 π β π£π 2 + ππ β ππ + 3ππ£ 2 + 6π£ 2 π = 0 The third equation is fairly easy to solve as solving for c gives , π = β15π£ 4 +3ππ£+π 2 3π£ now plugging into the fourth equation and cleaning up the expression, we get π β27π£ 6 +27ππ£ 3 +π 3 3π£ = 0 Multiplying both sides by , and dividing by we see that we get a similar 3π£ β 27 sextic π£ 6 β ππ£ 3 β ( π 3 ) 3 = 0 After solving for we see a similar expression. π£ π£ = 3 π 2 + ( π 2 ) 2 + ( π 3 ) 3 Choosing the positive square root from the quadratic formula we say that is already found. The issue is solving for π£ π’ Solving for π’ Looking at the and constant from our sextic earlier we see that we π’ 3 have π’ 6 + (10π£ 3 β 5π£π + β27π£ 6 +27ππ£ 3 +π 3 3π£ + π)π’ 3 + ( β27π£ 6 +27ππ£ 3 +π 3 3π£ )(π£ 3 + ππ£ + π) = 0 Solving the sextic for we get the following expression, π’ π’ = 3 β( 10π£ 3 β5π£π+ β 27π£ 6 +27ππ£ 3 +π 3 3π£ +π) 2 Β± ( ( 10π£ 3 β5π£π+ β 27π£ 6 +27ππ£ 3 + π 3 3π£ +π) 2 ) 2 β ( ( β27π£ 6 +27ππ£ 3 +π 3 3π£ )( π£ 3 +ππ£+π) 3 ) 3 Plugging in would give us a unwieldy formula in terms of . On top of π£ π, πππ π that, due to the in our solution for we end up with two solutions for , one of Β± π’ π₯ which will be incorrect. The reason we have 2 possible solutions for is because π₯ we multiplied the arbitrary cubic by a new cubic and doing so introduced a new solution for which may output an incorrect solution. π’ This is a picture of how the solution looks like written in LaTeX References History of cubic equations by David S. Richeson