Higher Algebra by Hall & Knight: Chapter 1, Art 16, Example 3

Question

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 $(5)$?

Answer 1

Hint: Try the other way around: expand the parentheses in (5).

Answer 2

Here is one way to see how the identity arises.

If we translate $a,b,c$ by $x$, the expression (without $k$) on the LHS of $(4)$ becomes $$ p(x)=(x+b)(x+c)(b-c)+(x+c)(x+a)(c-a)+(x+a)(x+b)(a-b). $$ However, $p$ is a constant polynomial, because the coefficient of $x$ is $(b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b)=0$, and the coefficient of $x^2$ is $(b-c)+(c-a)+(a-b)=0$. Therefore $p(0)=p(-a)$, i.e. $$ bc(b-c)+ca(c-a)+ab(a-b)=-(a-b)(c-a)(b-c). $$