A cube shifted is solved

BREAKING-CHANGES.md

@@ -116,6 +116,18 @@ did, and `ln(x + 2) / ln(x + 3)` at 0 was `ln(2) / ln(3)`. With symbols it was n
 | `"limit(ln(a + 1/x) / ln(b + 1/x), x, +oo)".ToEntity().Simplify()` | `a / b` | `ln(a) / ln(b)` |
 | `"limit(ln(1 + x) / ln(y + x), x, 0)".ToEntity().Simplify()` | `1 / y` | `0 provided not ln(y) = 0` |
 
+### A cube shifted is solved
+
+**Silent: a cubic `a(x + h)^3 = k` written out, with `b^2 = 3ac`, is solved where it was answered `{ }`.** Cardano's
+`C^3` is `(D1 + sqrt(D1^2 - 4 D0^3)) / 2`, and with `D0 = 0` that is `(D1 + |D1|) / 2`, which is 0 for a negative `D1`.
+Every root was then taken for `-b/(3a)`, which is none, and the empty set claimed there was no root. With `D0 = 0`, `C` is a
+cube root of `D1` itself.
+
+| Input | Was (2.6.0) | Now |
+|---|---|---|
+| `"(x + 1)^3 = 2".ToEntity().Solve("x")` | `{ }` | `{ -(3 + -3 * 2 ^ (1/3)) / 3, … }`, the three cube roots of 2 less 1 |
+| `"3x^3 + 9x^2 + 9x + 1 = 0".ToEntity().Solve("x")` | `{ }` | `{ -(9 + -3 * 18 ^ (1/3)) / 3 / 3, … }` |
+
 ### A quadratic whose discriminant is a square is answered without the root
 
 **Silent: the same roots, written as what they are.** The quadratic formula wrote `sqrt(b^2 - 4ac)` whatever the

Sources/.editorconfig

@@ -96,6 +96,9 @@ file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed
 [Tests/UnitTests/Convenience/EmptySequenceFoldTest.cs]
 file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed under MIT.\nDetails: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.\nWebsite: https://am.angouri.org.\n
 
+[Tests/UnitTests/Algebra/SolveTest/AShiftedCubeTest.cs]
+file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed under MIT.\nDetails: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.\nWebsite: https://am.angouri.org.\n
+
 [Tests/UnitTests/Convenience/FoldEnumerationTest.cs]
 file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed under MIT.\nDetails: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.\nWebsite: https://am.angouri.org.\n
 

Sources/AngouriMath/Functions/Continuous/Solvers/EquationSolver/PolynomialSolver.cs

@@ -239,10 +239,23 @@ internal static IEnumerable<Entity> SolveCubic(Entity a, Entity b, Entity c, Ent
             var u3 = Rational.Create(-1, 2) - coeff;
             var D0 = MathS.Sqr(b) - 3 * a * c;
             var D1 = (2 * MathS.Pow(b, 3) - 9 * a * b * c + 27 * MathS.Sqr(a) * d).InnerSimplified;
-            var C = MathS.Pow((D1 + MathS.Sqrt(MathS.Sqr(D1) - 4 * MathS.Pow(D0, 3))) / 2, Rational.Create(1, 3));
+            // C is a cube root of either root of C^6 - D1 C^3 + D0^3 = 0, and only one that is not 0
+            // can be divided by. With D0 = 0 those roots are D1 and 0, and the square root of D1^2
+            // is |D1|, so for a negative D1 the sum below is 0: a(x + h)^3 = k, a cube shifted, had
+            // C = 0 and every root taken for -b/(3a), which is none, and (x + 1)^3 = 2 was answered
+            // { }. With D0 = 0 a cube root of D1 itself is C, and the roots are -(b + u C)/(3a),
+            // D0/C being 0; that is -b/(3a) three times where D1 is 0 as well. For a negative D1 the
+            // cube root is written as the negative of the cube root of -D1, which is a cube root of
+            // D1 whichever root of a negative number a power is read as.
+            var shifted = D0.Evaled == 0;
+            var C = !shifted
+                ? MathS.Pow((D1 + MathS.Sqrt(MathS.Sqr(D1) - 4 * MathS.Pow(D0, 3))) / 2, Rational.Create(1, 3))
+                : D1.Evaled is Real { IsNegative: true }
+                    ? -MathS.Pow(-D1, Rational.Create(1, 3))
+                    : MathS.Pow(D1, Rational.Create(1, 3));
 
             return new[] { u1, u2, u3 }.Select(uk =>
-                C.Evaled == 0 && D0.Evaled == 0 ? -(b + uk * C) / 3 / a : -(b + uk * C + D0 / C / uk) / 3 / a);
+                shifted ? -(b + uk * C) / 3 / a : -(b + uk * C + D0 / C / uk) / 3 / a);
         }
 
         [ConstantField] private static readonly int[] sqrtsOf1 = { -1, 1 };

Sources/Tests/UnitTests/Algebra/SolveTest/AShiftedCubeTest.cs

@@ -0,0 +1,48 @@
+//
+// Copyright (c) 2019-2026 Angouri.
+// AngouriMath is licensed under MIT.
+// Details: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.
+// Website: https://am.angouri.org.
+//
+
+using AngouriMath.Extensions;
+using Xunit;
+using static AngouriMath.Entity;
+using static AngouriMath.Entity.Set;
+
+namespace AngouriMath.Tests.Algebra
+{
+    /// <summary>
+    /// A cube shifted, <c>a(x + h)^3 = k</c>, has <c>D0 = b^2 - 3ac = 0</c>, and Cardano's
+    /// <c>C^3 = (D1 + sqrt(D1^2 - 4 D0^3)) / 2</c> is then <c>(D1 + |D1|) / 2</c>, which is 0 for a
+    /// negative <c>D1</c>. Every root was taken for <c>-b/(3a)</c>, none was one, and
+    /// <c>(x + 1)^3 = 2</c> was answered <c>{ }</c>. Found by a cross-check of equations in
+    /// logarithms, one of which exponentiates to <c>3x^3 + 9x^2 + 9x + 1</c>; no issue was filed.
+    /// </summary>
+    [Trait("Area", "Algebra")]
+    public sealed class AShiftedCubeTest
+    {
+        [Theory]
+        [InlineData("(x + 1)^3 = 2", "-1", "2")]
+        [InlineData("x^3 + 3x^2 + 3x - 1 = 0", "-1", "2")]
+        [InlineData("3x^3 + 9x^2 + 9x + 1 = 0", "-1", "2/3")]
+        [InlineData("x^3 - 3x^2 + 3x - 3 = 0", "1", "2")]
+        // D1 positive, which was answered: (x + 1)^3 = -2.
+        [InlineData("x^3 + 3x^2 + 3x + 3 = 0", "-1", "-2")]
+        public void ItsRootsAreTheCubeRootsShifted(string equation, string shift, string cube)
+        {
+            var answer = Assert.IsType<FiniteSet>(equation.ToEntity().Solve("x"));
+            Assert.Equal(3, answer.Count);
+            foreach (var unity in new[] { "1", "-1/2 + sqrt(3) / 2 * i", "-1/2 - sqrt(3) / 2 * i" })
+            {
+                // The real cube root of the cube, times each cube root of unity, shifted.
+                var root = $"({shift}) + ({unity}) * sgn({cube}) * abs({cube})^(1/3)".ToEntity();
+                Assert.Contains(answer, found => Number.IsZero((found - root).EvalNumerical()));
+            }
+        }
+
+        [Fact]
+        public void ATripleRootIsStillOne()
+            => Assert.Equal("{ -1 }".ToEntity(), "x^3 + 3x^2 + 3x + 1 = 0".ToEntity().Solve("x"));
+    }
+}