Cubic formula

The cubic formula states that the roots of $x^3+ax+b=0$ are of the form

$$r=-A-B, -A\omega-B\omega^2, -A\omega^2-B\omega,$$

where $\omega= -{1\over 2}+{\sqrt{3}i\over 2}$, and

$$A=({b\over 2}+({b^2\over 4}+{a^3\over 27})^{1/2})^{1/3},$$

$$B=({b\over 2}-({b^2\over 4}+{a^3\over 27})^{1/2})^{1/3},$$

Any cubic polynomial $y^3+py^2+qy+r$ can be reduced to the form above using the substitution $y=x-{p\over 3}$, where

$$a=q-{p^2\over 3},\ \ \ \ \ b={2p^3-9pq+27r\over 27}.$$

There is a similar (though much more complicated) general formula for the roots of quartic polynomials.

Exercise 2.10.4   Derive the above formula for the solution of the cubic. [Hint: One approach is the following. Make the substitution $x=m+n$, so $(m+n)^3+a(m+n)+b=0$. If $m^3+n^3+b=0$ and $3mn+a=0$ then $m^3+3mn(m+n)+n^3+a(m+n)+b=0$. Solve for $m$, $n$ in terms of $a$, $b$.