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

```
9 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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Sagar Singh
B.Tech, IIT Delhi

```
9 years ago
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