Dr. Math FAQ ||
Classic Problems ||
Formulas ||
Search Dr. Math ||
Dr. Math Home

## Another Solution of the Cubic EquationHere is another method of solving a cubic polynomial equation submitted independently by Paul A. Torres and Robert A. Warren. It is based on the idea of "completing the cube," by arranging matters so that three of the four terms are three of the four terms of a perfect cube.Start with the cubic equation If ax ^{2} - 3b = 0, then the first three terms are the first three
terms of a perfect cube, namely (x+a/3)^{3}. Then you can "complete
the cube" by subtracting c from both sides and adding the missing term of
the cube a^{3}/27 to both sides. Recalling that b = a^{2}/3,
you get
By taking the cube root of the left side and the three cube roots of the right side, you getx These are the roots of the cubic equation that were sought.x = -a/3 + (a
If a where(y+z) The first three terms of this equation in y will be those of a perfect cube if and only if dd = 3z + a, e = 3z ^{2} - 3e is zero, which happens if and only if
a^{2} - 3b = 0, which cannot happen in this case, so we seemingly
haven't gained anything. However, the last three terms of this
equation in y will be those of a perfect cube if and only if e^{2} = 3df,
that is if and only if
where(3z Since ag = a ^{2} - 3b is nonzero, g is nonzero, and we have a true quadratic
equation, called the resolvent quadratic. Now we pick z to be a root of this
quadratic equation. If the GCD of z^{2} + (h/g)z + (i/g) and
z^{3} + az^{2} + bz + c as polynomials in z is not 1, then
any root of the GCD is also a root of the original cubic equation in x. Once
you have at least one root, the problem of finding the other roots is reduced to
solving a quadratic or linear equation. If the GCD is 1, then neither value of
z can make f = 0, so we can assume henceforth that f is nonzero. Either root
z of the quadratic will do, but we must choose one of them. We arbitrarily
pick the one with a plus sign in front of the radical,
Set z equal to this value in the equation for y, and divide it by f on both sides. Then the last three terms of the cubic in y are those of a perfect cube, namely (ey/[3f]+1)z = (-h+sqrt[h ^{3}, so we can complete the cube to solve it.
We do this by subtracting y^{3}/f from both sides, then adding the
missing term of the cubic, (ey/[3f])^{3} to both sides, obtaining
Now you have the values of y. Add z to each to get the values of x:(ey/[3f]+1) These are the roots of the cubic equation that were sought.x = z + 3f/[(e Example: Solve x^{3} + 6x^{2} + 9x + 6 = 0.
Thena = 6, b = 9, c = 6. Thend = 3z + 6, e = 3z and the resolvent quadratic isg = a Then9z and the cubic in y isd = 6 + 3 sqrt(3), e = 18 + 12 sqrt(3), f = 24 + 12 sqrt(3), Then one root isy After a lot of simplification, you gety = 3f/[(e (and two other roots).y = -2 - sqrt(3) - (2-sqrt[3]) |

[**Privacy Policy**]
[**Terms of Use**]

Math Forum Home ||
Math Library ||
Quick Reference ||
Math Forum Search

© 1994-2015 Drexel University. All rights reserved.

http://mathforum.org/

The Math Forum is a research and educational enterprise of the Drexel University School of Education.