## Friday, November 24, 2006

### Depressed Cubic

Scipione del Ferro's great discovery which eventually led to the general solution of the cubic equation was the solution of the depressed cubic equation:

x3 + bx = c

In today's blog, I will show how del Ferro solved this equation. The content is taken from Paul J. Nahin's The Story of √-1

(1) Let x = u+v

(2) Then:

x3 + bx = (u + v)3 + b(u + v) =

= u3 + 3u2v + 3uv2 + v3 + b(u + v) =

= u3 + v3 + (3uv + b)(u + v) = c

(3) Now, if:

(3uv + b) = 0

then:

u3 + v3 = c

(4) Now,

(3uv + b) = 0 → 3uv = -b

and further:

v = -b/(3u)

(5) So if:

u3 + v3 = c, then:

u3 + [-b/(3u)]3 - c =

u3 + (-b3)/(27u3) - c =

u6 + (-b3/27 - cu3 = 0

(6) But this is an example of a quadratic equation since if:

z = u3, then we have:

u6 + (-b3/27) - cu3 =

= z2 - cz -(b3/27)

(7) Using the solution for the quadratic equation (see here), we have:

z = (c/2) ± √c2 - 4(-b3/27)/2 =

=(c/2) ± √c2/4 + b3/27 = u3

(8) Considering only the positive form, we have:

(9) Since v3 = c - u3, we have:

v3 = c - (c/2 + √c2/4 + b3/27 =

= c/2 - √c2/4 + b3/27

(10) This then gives us:

(11) Since x = u + v, this gives us that:

(12) Since (-1)3 = (-1)

(13) This tells us that we can restate (11) as the following:

In another blog, I present a proof that this really is the solution of the depressed cubic.

References

Ramanathan Somasundaram said...

Hi there,

I found your explanation very useful for solving cubic equation. Up to step 11, I followed it well. But in Step 12, I could not understand why -1 (cube) calculation happened. Also, my intent was to find out the roots for a cubic equation. If you could add some details on finding the roots for a cubic equation (not just the depressed cubic), it will be very helpful - Ram from Bangalore, India

Ramanathan Somasundaram said...

One additional point, what I am trying to figure out is; why should I restate 11 and 13. Now, I followed the step 12 (i.e.) 1 was termed as -1 * -1 and then -1 was brought into the cubic root. and thus 11 was equated with 13. But how does the equation help. What purpose does it serve?fvu

Larry Freeman said...

Hi Ramanathan,

Thanks very much for your question.

As you point out accurately, the two equations step #11 and step #13 are equivalent.

I present both steps just to show that they are equal to each other.

In books about the depressed cubic, they might show the equation in #11 or the equation in #12.

If you are interested in seeing the full proof for the cubic equation, I have a blog on it here.

Cheers,

-Larry

bharti said...

i want to find the solution of the eq 8x^3-6x+=0

bharti said...

how to solve the cubic equations of form Ax^3-Bx+C=0

bharti said...

i ve been trying to solve the eq 8x^3-6x+1=0 but havent got any success so far

auto-df said...

Great, a much clearer method to follow than in Nahins book. Thanks for taking the time to do this.