 # the procedure for the problem whose cubic equation is given and to find out the real root is there in the interval.

12 years ago

Dear student,

The general cubic equation is

`         A x3  + B x2  + C x + D = 0`

The coefficients A, B, C, D are real or complex numbers with A not 0. Dividing through by A, the equation comes to the form

`          x3  + b x2  + c x + d = 0`

Now we want to reduce the last equation by the substitution

`         x = y + r`

The cubic equation becomes:

`         (y + r)3  + b (y + r)2  + c (y + r) + d = 0<=>        y3  + (3 r + b) y2  + (3 r2  + 2 r b + c) y + r3  + r2  b + r c + d = 0`

Now we choose y such that the quadratic term disappears

`         choose  r = -b/3So, with te substitution                b        x = y - -                3the equation         x3  + b x2  + c x + d = 0comes in the form        y3  + e y + f = 0`

## Vieta's substitution

tr vieta-substitution To reduce the last equation we use the Vieta subtitition

`                   1        y = z + s -                  z`

The constant s is an undefined constant for the present.
The equation

`         y3  + e y + f = 0becomes             s        (z + -)3  + e (z + (s/z)) + f = 0             zexpanding an multiplying through by z3  , we have        z6  + (3 s + e) z4  + f z3  + s (3 s + e) z2  + s3  = 0`

Now we choose s = -e/3.
The equation becomes

`         z6  + f z3  - e3/27 = 0With    z3  = u        u2  + f u -e3/27 = 0`

This is an easy to solve quadratic equation.

All the best.

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