cubic equation

[Guest]

x^3+3ax^2+3ax-a^3=0
How to solve this in terms of X
a is bigger than 0 and positive.
I do not know how to start and how to proceed.
Thanks in advance

[Guest]

https://www.math10.com/en/algebra/cubic ... tions.html

[Guest]

Are you sure you are not missing an "a" in that? If it were "[tex]a^2x[/tex]" rather than "[tex]ax[/tex]" i5 would be easy!

[tex](x- a)^3= x^3- 3ax^2+ 3a^2x- a^3[/tex] so that would be [tex](x- a)^3= 0[/tex] and x= a is a triple root.

[Guest]

There is a "cubic formula", much more complicated than the "quadratic formula". Here is my take on it:
For any numbers, a and b, [tex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/tex] and [tex]3ab(a- b)= 3a^2b- 3ab^2[/tex] so [tex](a- b)^3+ 3ab(a- b)= a^3- b^3[/tex].

If we let x= a- b, p= 3ab, and [tex]q= a^3- b^3[/tex] that becomes the "reduced cubic" [tex]x^3+ px= q[/tex].
("Reduced" because there is no [tex]x^2[/tex] term. Given any cubic equation, it is always possible to change the variable to "reduce" the equation.)

So, if we are given a and b we can construct a cubic equation having a- b as a root. What about the other way? Given p and q can we find a and b and so x= a- b?

Yes, we can! From p= 3ab, b= p/3a. Then [tex]q= a^3- b^3= a^3- p^3/(3^3a^3)[/tex] so, multiplying by [tex]a^3[/tex], [tex]qa^3= a^6- \frac{p^3}{3^3}[/tex] or [tex]a^6- qa^3- \left(\frac{p}{3}\right)^3= 0[/tex].

Now that is a sixth-degree equation, which could be much worse than a cubic, but since the only powers are multiples of 3, we can let [tex]y= a^3[/tex] and have the quadratic equation [tex]y^2- qy- \left(\frac{p}{q}\right)^3= 0[/tex].

Using the quadratic formula, [tex]y= a^3= \frac{q\pm\sqrt{q^2+ 4\left(\frac{p}{q}\right)^3}}{2}[/tex].

We have a choice of "+" or "-" in the quadratic formula for y and then y will have 3 roots for a so a possible 6 roots, but there are only 3 distinct roots for a. Once we have a, [tex]b= \frac{p}{3a}[/tex] and x= a- b.

[Guest]

Very well explained!
Thanks.