This is a solved example from Higher Algebra by Hall and Knight. There are a few steps missing so I can't understand the continuity of the solution.
Chapter 1, Ratio, Art 16, Example 3
Solve the Equations $$ax + by + cz = 0, \tag{1}\label{eq1}$$
$$x + y + z = 0, \tag{2}\label{eq2}$$
$$bcx +cay + abz = (b-c) (c-a) (a-b), \tag{3}\label{eq3}$$
From (1) and (2) by cross multiplication,
$$\frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} = k,\ \text {suppose;}$$
$$\therefore x = k(b-c), y = k(c-a), z = k(a-b)$$.
Substituting in (3),
$$k \ [bc(b-c) + ca(c-a) + ab(a-b)] = (b-c)(c-a)(a-b), \tag{4}\label{eq4}$$
[OP's Note: I have understood till here, and I can't figure out how the authors derived the equation given in the next line.]
$$k \ [-(b-c)(c-a)(a-b)] = (b-c)(c-a)(a-b), \tag{5}\label{eq5}$$
$$\therefore \ k=-1$$
whence: $$x=c-b,\ y=a-c,\ z=b-a$$
How is $(4)$ reduced to $(5)$?