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Mathematics Department Solving a Cubic Equation The quadratic formula for the general degree two equation is one of the most familiar equations in mathematics. A formula exists for solving the general cubic equation but it is not easily quotable. I was surprised at discovering a non-standard derivation of the formula in an article in the library whilst I was looking for something else. It goes as follows: The solutions of x 3 px q 0 are x (b c) 2 x (b c), x (b c) 1 i 3 q q 2 (4 p 3 27) where 3 1 and and b 3 and c p 3b . 2 2 Proof We wish to solve the cubic ax 3 bx 2 cx d 0 . Using a linear change of variable, this cubic can be reduced to: x 3 px q 0 . (1) Now the following identity holds (a, b and c are arbitrary complex constants): (a b c)(a b c)(a b c) (a 3 3abc b 3 c 3 ) Replacing a with x, this produces: (x b c)( x b c)( x b c) ( x 3 3xbc b 3 c 3 ) (2) Now we can recognise that the right side of (2) is essentially the same as what appears in (1), provided the following hold: 3bc p (3) b c q 3 3 (4) So given the values of p and q, we need only determine a b and c satisfying these relations. (2) will then give three linear factors and hence three solutions. We can rewrite the equations (3) and (4) as: p3 b 3c 3 27 b c q 3 3 3 One can see that b 3 and c 3 are roots of the quadratic equation x 2 qx p 0. 27 1 q q 2 (4 p 3 27) 3 . Let b 3 q q (4 p 27) 2 3 These roots are 2 2 and by (3) c p 3b . (b c) Hence the solutions of (1) are x , x (b c), x (b c) JL/WGN 24 May, 2011

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cubic equation, the roots, quadratic equation, three roots, complex numbers, how to, quartic equations, cube roots, square root

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posted: | 5/24/2011 |

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