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So, You Want to Solve the Cubic? Dana P. Williams 9 April 2009 Dana P. Williams So, You Want to Solve the Cubic? First Simplification We want to solve the equation y 3 + by 2 + cy + d = 0. (1) This, as we shall see, is not so easy. Therefore, we make a simplification that has been around for centuries. We let b x := y + . 3 Then, solving for x, we get x 3 − 3px + 2q = 0. This is the equation we’ll concentrate on now. Dana P. Williams So, You Want to Solve the Cubic? (2) A little Calculus Lemma Consider equation (2) (that is, x 3 − 3px + 2q = 0) with p 6= 0. (Equation (2) is pretty easy to solve if p = 0.) Then let D := q2 − p3 . 1 If D > 0, then (2) has exactly one real root. 2 If D = 0, then (2) has exactly two distinct real roots — one of them is repeated. 3 If D < 0, then (2) has three distinct real roots. Proof. This result is proved using a little one variable calculus. I am going to leave it as a bit of a challenge. Do it! Dana P. Williams So, You Want to Solve the Cubic? The Next Ancient Trick To solve (2), we introduce an extra variable: x = u + v. Then plugging into (2), we get (u + v )3 − 3p(u + v ) + 2q = 0. A little algebra brings us to u 3 + v 3 + (u + v )(3uv − 3p) + 2q = 0. (3) Since we introduced an extra variable, we can impose an additional requirement. Namely, uv = p. Using this condition in (3) gives us the simple equation u 3 + v 3 = −2q. Dana P. Williams So, You Want to Solve the Cubic? We Need Another Equation Involving u and v Since u 3 + v 3 = −2q, we can square both sides to get u 6 + 2u 3 v 3 + v 6 = 4q 2 . Then we get (u 3 − v 3 )2 = u 6 − 2u 3 v 3 + v 6 = u 6 + 2u 3 v 3 + v 6 − 4u 3 v 3 which, in view of (4) and uv = p, is = 4q 2 − 4p 3 = 4D. 1 Thus if D 2 is some choice of square root for D, then 1 u 3 − v 3 = 2D 2 . Dana P. Williams So, You Want to Solve the Cubic? (4) Solving for u and v Now we have two equations: ( u3 + v 3 u3 − v 3 = −2q 1 = 2D 2 . (5) Now some easy algebra gives 1 u 3 = −q + D 2 1 and v 3 = −q − D 2 . 1 (6) Note that a different choice for D 2 just interchanges the formulas for u 3 and v 3 . Therefore u + v , uv and u 3 + v 3 + 2q don’t care which choice we make. This just means that we are free to choose For example, when D > 0, the natural something convenient. √ choice is simply D. Dana P. Williams So, You Want to Solve the Cubic? The Case D > 0 Theorem If D > 0, then the unique real root of (2) is q q √ √ 3 3 x1 = −q + D + −q − D. The remaining (complex roots) can be found the quadratic q using √ 3 formula (after division), or by letting u1 = −q + D and q √ 3 v1 = −q − D. Then the complex roots are √ u1 + v1 3 x2 = − +i (u1 − v1 ) and 2 2 Dana P. Williams x3 = x̄2 . So, You Want to Solve the Cubic? The Proof of the First Theorem Proof. Let ω := − 21 + i √ √ 3 2 . Thus the cube roots of 1 are 1, ω and = − i 23 . Now solutions: x = uk + vj ω2 − 12 uk := ω k−1 taking cube roots in (6) gives 9 possible (for k, j = 1, 2, 3) where q q √ √ 3 j−1 3 −q + D and vj := ω −q − D . But these give solutions to (2) only when p p = uk vj = ω k+j−2 3 q 2 − D = ω k+j−2 p. Thus the solutions are x1 = u1 + v1 , x2 = u2 + v3 and x3 = u3 + v2 . This clearly gives the formula claimed for x1 . I’ll leave the algebra for x2 and x3 to you. Dana P. Williams So, You Want to Solve the Cubic? An Example Example (Just how great is this?) Consider the equation x 3 − 3x − 18 = 0. In this example, p = 1 and q = −9. Since D = q 2 − p 3 = 80 > 0, there is a unique real root. Namely q q √ √ 3 3 x1 = 9 + 80 + 9 − 80. This is all very exciting until you realize that x1 = 3. For example, √ √ √ √ (3 + 5)3 72 + 32 5 9 + 80 = 9 + 4 5 = = . 8 8 Thus u1 = √ q √ 3+ 5 3 9 + 80 = . 2 Similarly: v1 = √ 3− 5 . 2 Thus u1 + v1 = 3. You can also check that x2 = − 32 + i x3 = − 32 − √ i 215 . Dana P. Williams √ 15 2 So, You Want to Solve the Cubic? and The Case D = 0 Theorem If D = 0, then the two distinct real roots of (2) can be computed √ √ as follows. If q > 0, then the roots are −2 p and p. If q < 0, √ √ then the roots are 2 p and − p. Proof. √ Since D = 0, we have q = ±p p. Now just plug in and check. Dana P. Williams So, You Want to Solve the Cubic? The Cool Case: D < 0 Notice that if D = q 2 − p 3 < 0, then p > 0. Theorem Suppose that D < 0. Let θ := cos−1 −q √ . p p Then the three distinct real roots of (2) are given by θ + 2πk √ xk = 2 p cos k = 0, 1, 2. 3 Dana P. Williams So, You Want to Solve the Cubic? The Proof √ √ 1 We choose D√2 = i −D. Then u 3 = −q + i −D and v 3 = −q − i −D. p p √ Note that |u|3 = |u 3 | = q 2 + D = p 3 = p p. √ On the other hand, Im(u 3 ) > 0, so θ := cos−1 p−q p is an argument for u 3 . √ Thus the solutions tou 3 = −q + i −D are given by √ uk = p exp i θ+2πk for k = 0, 1, 2. 3 Since Im(v 3 ) < 0, a good choice for the argument of v 3 is −θ, where θ is the argument foru 3 . √ Thus vj = p exp i −θ+2πj for j = 0, 1, 2. 3 Dana P. Williams So, You Want to Solve the Cubic? More of the Proof Recall that we require that 2π(k + j) p = uk vj = p exp i . 3 Therefore our solutions should be u0 + v0 , u1 + v2 and u2 + v1 . Then, for example, u0 + v0 = θ θ √ θ √ p exp i + exp −i = 2 p cos . 3 3 3 A bit more work is required to see that u1 + v2 and u2 + v1 given the formulas in the statement of the theorem. Dana P. Williams So, You Want to Solve the Cubic? An Example with Three Real Roots Example Consider x 3 − 9x − 9 = 0. Here p = 3 and q = − 92 . Then D = 81 0. So we plug into the formula in the theorem to 4 − 27 <√ get θ = cos−1 23 = π6 . So are solutions are √ π ≈ 3.4115 2 3 cos 18 √ 13π 2 3 cos ≈ −2.2267 18 √ 25π 2 3 cos ≈ −1.1848. 18 Dana P. Williams So, You Want to Solve the Cubic? Just Checking 15 10 5 K3 K2 K1 0 K5 1 2 3 4 x K10 K15 Figure: The graph of f (x) = x 3 − 9x − 9. Dana P. Williams So, You Want to Solve the Cubic?