home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!charon.amdahl.com!pacbell.com!sgiblab!zaphod.mps.ohio-state.edu!sol.ctr.columbia.edu!The-Star.honeywell.com!umn.edu!uum1!ncrtimes!csdpc231.stpaul.ncr.com!preece
- From: preece@StPaul.c10sd4.NCR.COM (Bently.Preece)
- Subject: Re: Solutions to a cubic equation
- Message-ID: <preece.3@StPaul.c10sd4.NCR.COM>
- Sender: news@NCRTimes.stpaul.ncr.com
- Organization: NCR NPD, St. Paul, MN
- References: <1992Nov16.221527.0313263@locus.com> <17NOV199211491821@csa3.lbl.gov>
- Distribution: usa
- Date: Tue, 17 Nov 1992 23:29:46 GMT
- Lines: 23
-
-
- markd@locus.com (Mark Dubinsky) writes
-
- ... What is the formula for the solutions of a cubic
- equation
-
- x3 + ax2 + bx + c = 0 ?
-
- sichase@csa3.lbl.gov (SCOTT I CHASE) writes:
-
-
- >Eliminate the Y^2 term by suitable substitution.
-
- X^3 + D*X + E = 0 (**)
-
- ... The idea is to
- use the trignometric identity:
-
- 4*cos^3(x) - 3*cos(x) - cos(3*x) = 0. (***)
-
- Forget trig identities. Try (A + B)^3 = 3AB(A + B) + (A^3 + B^3). Now if
- you can get 3AB = -D and (A^3 + B^3) = -E then A + B will be a solution.
- This is easy to solve.
-