Solving a cubic equation

This video is about a cubic equation
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[louisromao]

Try this approach:

(x-1)^3 - (x·7^(1/3))^3 = 0

Use the difference of cubes formula to come up with the real solution (which is the same as what you derived), and then use the quadratic formula to come up with the two imaginary solutions.

[maheshbhatt]

, Namaste Sir .As this is an odd degree ( 3 ) equation it has at least one real root. Now substituting 0, 1, 2 ...etc for x it is clear that root is not positive .Then putting x = -1 , -2 .., and using De Carte rule of sign the root lies between -1 and -2.The exact solution also confirms this.

[matthiasbergner]

4:20 : a^3 = b^3 only implies a = b in the real numbers, but not in the complex numbers.

[holyshit]

First step
Get rid of x^2 term (You can do this by expressing cubic as sum of powers of the binomial, fe using Horner's scheme repeatedly)
Second step
Use substitution
y=u+v // Developed by Fontana vel Tartaglia (it is not known how del Ferro did it)
y=p/(3z)-z // Developed by Francois Viete (yes Harriot is also credited but Harriot lived a little bit later)
You should get quadratic equation in z^3 after using Vieta substitution
If you use Fontana substitution you should rewrite equation as system of two equations and then transform this system of equations
to the Vieta formulas for quadratic
I prefer Fontana substitution because in Vieta substitution it is possible to get division by zero

[JoePortly]

I do like the elegance and simplicity of this solution. And getting there reminds me of a question about cubic roots. As those of a square may be expressed as ± n^½, how should n^(1/3) be stated? Oughtn't the expression comprise several possibilities, subject to the positivity or negativity of n? Could the cubic root sign be introduced or preceded by, say, +++/+ - - or - - -/- + +, or is this question irrelevant?