Cubic equation

In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. An example is the equation

Lagrange resolvents

The symmetric group S3 of order three has the cyclic group of order three as a normal subgroup, which suggests making use of the discrete Fourier transform of the roots, an idea due to

Related Topics:
Symmetric group - Cyclic group - Normal subgroup - Discrete Fourier transform

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Lagrange.

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Suppose that r0, r1 and r2 are the roots of equation (1), and define zeta = (-1+isqrt{3})/2, so that ζ is a primitive third root of unity. We now set

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:s_0 = r_0 + r_1 + r_2,,

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:s_1 = r_0 + zeta r_1 + zeta^2 r_2,,

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:s_2 = r_0 + zeta^2 r_1 + zeta r_2.,

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The roots may then be recovered from the three si by inverting the above linear transformation, giving

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:r_0 = (s_0 + s_1 + s_2)/3,,

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:r_1 = (s_0 + zeta^2 s_1 + zeta s_2)/3,,

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:r_2 = (s_0 + zeta s_1 + zeta^2 s_2)/3.,

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We already know the value s0 = −a, so we only need to seek values for the other two. However, if we take the cubes, a cyclic permutation leaves the cubes invariant, and a transposition of two roots exchanges s13 and s23, hence the polynomial

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:(z-s_1^3)(z-s_2^3) qquad (5)

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is invariant under permutations of the roots, and so has coefficients expressible in terms of (1). Using calculations involving symmetric functions or alternatively field extensions, we can calculate (5) to be

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:{z}^{2}+ left( -9,ba+2,{a}^{3}+27,c ight) z+ left( {a}^{2}-3,b ight)^{3}.

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The roots of this quadratic equation are

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:rac92,ab-{a}^{3}- rac{27}{2},c pm rac32,sqrt{3Delta},

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where Δ is the discriminant defined above. Taking cube roots give us s1 and s2, from which we can recover the roots ri of (1).

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