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

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

The quadratic term disappears

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/3

So, with te substitution

                b
        x = y - -
                3
the equation

         x3  + b x2  + c x + d = 0

comes 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 = 0
becomes

             s
        (z + -)3  + e (z + (s/z)) + f = 0
             z

expanding 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 = 0

With    z3  = u

        u2  + f u -e3/27 = 0

This is an easy to solve quadratic equation.

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi