Euler’s derivation of the quartic formula

Let us consider the quartic equation

$y^4+p{y}^2+qy+r=0,$(1)

where $p,q,r$ are arbitrary known complex numbers.  We substitute in the equation

$y:=u+v+w.$(2)

We get firstly
$y^2=(u^2+v^2+w^2)+2(vw+wu+uv),$
$y^4={(u^2+v^2+w^2)}^2+4(u^2+v^2+w^2)(vw+wu+uv)+4(v^2{w}^2+w^2{u}^2+u^2{v}^2)+8uvw(u+v+w).$

Thus (1) attains the form

$$4(v^2{w}^2+w^2{u}^2+u^2{v}^2)+{(u^2+v^2+w^2)}^2+p(u^2+v^2+w^2)+r\mathit{\hspace{1em}\hspace{1em}\hspace{0.25em}}$$

$$+(vw+wu+uv)[4(u^2+v^2+w^2)+2p]+(u+v+w)[8uvw+q]=0.$$

When $u,v,w$ are determined so that

$u^2+v^2+w^2=-{\displaystyle \frac{p}{2}},$(3)

$uvw=-{\displaystyle \frac{q}{8}},$(4)

the expressions in the brackets vanish and our equation shrinks to the form

$v^2{w}^2+w^2{u}^2+u^2{v}^2={\displaystyle \frac{p^2-4r}{16}}.$(5)

Squaring (4) gives

$u^2{v}^2{w}^2={\displaystyle \frac{q^2}{64}}.$(6)

The left hand sides of (3), (5) and (6) are the elementary symmetric polynomials of $u^2$, $v^2$, $w^2$, whence these three squares are the roots $z_1$, $z_2$, $z_3$ of the so-called cubic resolvent equation

$z^3+{\displaystyle \frac{p}{2}}{z}^2+{\displaystyle \frac{p^2-4r}{16}}z-{\displaystyle \frac{q^2}{64}}=0.$(7)

Therefore we may write

$$u=\pm \sqrt{z_1},v=\pm \sqrt{z_2},w=\pm \sqrt{z_3}.$$

All 8 sign combinations of those square roots satisfy the equations (3), (5), (6). In order to satisfy also (4) the signs must be chosen suitably.  If  $u_0,v_0,w_0$ is some suitable combination of the values of the square roots, then all possible combinations are

$$u_0,v_0,w_0;u_0,-v_0,-w_0;-u_0,v_0,-w_0;-u_0,-v_0,w_0.$$

Accordingly, we have the

Theorem (Euler 1739).  The roots of the equation (1) are

$\{\begin{array}{cc}{y}_1=u_0+v_0+w_0,\hfill & \\ y_2=u_0-v_0-w_0,\hfill & \\ y_3=-u_0+v_0-w_0,\hfill & \\ y_4=-u_0-v_0+w_0,\hfill & \end{array}$(8)

where $u_0,v_0,w_0$ are square roots of the roots of the cubic resolvent (7).  The signs of the square roots must be chosen such that

$$u_0{v}_0{w}_0=-\frac{q}{8}.$$

The equations (8) imply an important formula

$(y_1-y_2)(y_1-y_3)(y_1-y_4)(y_2-y_3)(y_2-y_4)(y_3-y_4)=$$-2^6(v_0^2-w_0^2)(w_0^2-u_0^2)(u_0^2-v_0^2)$

$=$$-64(z_2-z_3)(z_3-z_1)(z_1-z_2),$

which yields the

Corollary.  A quartic equation has a multiple root always and only when its cubic resolvent has such one.