Solving Cubic when There are Known to be 3 Real Roots When solving for roots to a cubic equation, the sign of the $\Delta$ tells us when there will be 3 distinct real roots (as long as the first terms coefficient, $a$, is non-zero.) Namely when $\Delta$ is positive.
The equations to find the 3 roots are:


*

*$x_1 = -\frac{1}{3a}(b + C + \frac{\Delta_0}{C})$

*$x_2 = -\frac{1}{3a}(b + \frac{C(-1 + i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 + i\sqrt 3)})$

*$x_3 = -\frac{1}{3a}(b + \frac{C(-1 - i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 - i\sqrt 3)})$


Where: $$C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}}$$
By the given equation, $\Delta_1^2 - 4\Delta_0^3 = -27a^2\Delta$, I know that when $\Delta$ is positive the square root in $C$ will produce an $i$. So when $\Delta$ is positive $C$ is effectively: $$C = \sqrt[3]{\frac{\Delta_1 + i\sqrt{4\Delta_0^3 - \Delta_1^2}}{2}}$$
So obviously the $i$ in $C$ cancels with the $i$s in the $x_2$ and $x_3$ roots, and we get 3 real roots. But for the life of me I cannot work out how. Can someone help me break those steps down?

Given $x^3+a x^2+b x+c x+d=0$, substitute $x=y-a/3. $ This gives  $y^3+p y+q=0$ where $p,q$ are determined by $a,b,c,d$ and the substitution. Now let $y=u+v$. We have $$0=(u+v)^3+p(u+v)+q=(u^3+v^3+q)+(u+v)(3u v+p).$$ Suppose  there are non-zero $u,v$ such that $$u^3+v^3+q=0=3u v+p.$$  Substituting $v=-p/3u$ into $0=u^3+v^3+q$ and letting $w=u^3$, we have $$w^2+q w+(-p^3/27)=u^3(u^3+q+v^3)=0.$$ Solve the quadratic equation in $w . $  Let $u$ be any complex cube root of (either value of) $w$. Then $x=y-a/3=u+v-a/3=u-p/3u-a/3$ solves the original cubic....When there are 3 real solutions for $x$ and they are not an arithmetic sequence, the values of $w$ will not be real numbers and you need the cube root of one of them. One way is to put $w=r. cis (t)=r(\cos t+i\sin t)$ with $r>0$ and $t\in (0,2\pi). $ The cube roots of $w$ are $r^{1/3}cis ((t+2\pi  n)/3)$ for $n\in \{0,1,2\}.$.....The main idea in this method of solving the cubic  equationis to change the variable to a sum $u+v$ and see if the equation can be split into 2 simpler equations.
