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Theory of Equation
 Cubic Equations
 Sources and References

Theory of Equation

Cubic Equations

483 To solve the general cubic equation π‘₯3+𝑝π‘₯2+π‘žπ‘₯+π‘Ÿ=0 Remove the term 𝑝π‘₯2 by the method of (429). Let the transformed equation be π‘₯3+π‘žπ‘₯+π‘Ÿ=0 484 Cardan's method: The complete theoretical solution of this equation by Cardan's method is as follows:- Put π‘₯=𝑦+𝑧i. 𝑦3+𝑧3+(3𝑦𝑧+π‘ž)(𝑦+𝑧)+π‘Ÿ=0 Put 3𝑦𝑧+π‘ž=0; βˆ΄π‘¦=βˆ’π‘ž3𝑧 Substitute this value of 𝑦, and solve the resulting quadratic in 𝑦3. The roots are equal to 𝑦3 and 𝑧3 respectively; and we have, by [i] 485 π‘₯=βˆ’π‘Ÿ2+π‘Ÿ24+π‘ž32713+βˆ’π‘Ÿ2βˆ’π‘Ÿ24+π‘ž32713 The cubic must have one real root at least, by (409). Let π‘š be one of the three values of βˆ’π‘Ÿ2+π‘Ÿ24+π‘ž32713, and 𝑛 one of the three values βˆ’π‘Ÿ2βˆ’π‘Ÿ24+π‘ž32713. 486 Let 1, 𝛼, 𝛼2 be the three cube roots of unity, so that 𝛼=βˆ’12+12-3, and 𝛼2=βˆ’12βˆ’12-3472 487 Then, since 3π‘š3=π‘š31, the roots of the cubic will be π‘š+𝑛, π›Όπ‘š+𝛼2𝑛, 𝛼2π‘š+𝛼𝑛, Now, if in the expansion of βˆ’π‘Ÿ2Β±π‘Ÿ24+π‘ž32713 by the Binomial Theorem, we put πœ‡= the sum of the odd terms, and 𝜈= the sum of the even terms then we shall have π‘š=πœ‡+𝜈, and 𝑛=πœ‡βˆ’πœˆ; or else π‘š=πœ‡+πœˆβˆ’1, and 𝑛=πœ‡βˆ’πœˆβˆ’1; according as π‘Ÿ24+π‘ž327 is real or imaginary. 488 By substituting these expressions for π‘š and 𝑛 in (487), it appears that (i.) If π‘Ÿ24+π‘ž327 be positive, the roots of the cubic will be 2πœ‡, βˆ’πœ‡+πœˆβˆ’3, βˆ’πœ‡βˆ’πœˆβˆ’3 (ii.) If π‘Ÿ24+π‘ž327 be negative, the root will be 2πœ‡, βˆ’πœ‡+𝜈3, βˆ’πœ‡βˆ’πœˆ3 (iii.) If π‘Ÿ24+π‘ž327=0, the roots are 2π‘š, βˆ’π‘š, βˆ’π‘š since π‘š is now equal to πœ‡. 489 The Trigonometrical method: The equation π‘₯3+𝑝π‘₯2+π‘žπ‘₯+π‘Ÿ=0 may be solved in the following manner by Trigonometry, when π‘Ÿ24+π‘ž327 is negative.
Assume π‘₯=𝑛cos 𝛼. Divide the equation by 𝑛3; thus cos3𝛼+π‘žπ‘›2cos 𝛼+π‘Ÿπ‘›3=0 But cos3π›Όβˆ’34cos π›Όβˆ’cos 3𝛼4=0By (657) Equate coefficients in the two equations; the result is 𝑛=βˆ’4π‘ž312, cos 3𝛼=βˆ’4π‘Ÿβˆ’34π‘ž12 𝛼 must now be found with the aid of the Trigonometrical tables. 490 The roots of the cubic will be 𝑛cos 𝛼, 𝑛cos(23πœ‹+𝛼), 𝑛cos(23πœ‹βˆ’π›Ό) 491 Observe that, according as π‘Ÿ24+π‘ž327 is positive or negative, Cardan's method or the Trigonometrical will be practicable. In the former case, there will be one real and two imaginary roots; in the latter case, three real roots.

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science & Engineering
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions
  3. Stroud, K.A., 2001, Engineering Mathematics
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem
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