1
$\begingroup$

I am stuck with the following problem that says:

If $x(b-c)+y(c-a)+z(a-b)=0$ then show that $$\frac{bz-cy}{b-c}=\frac{cx-az}{c-a}=\frac{ay-bx}{a-b}$$ where $a \neq b \neq c.$

Can someone point me in the right direction? Thanks in advance for your time .

  • 1
    Please post the *complete* question. As it is now, it's false - take for example $a=b=c\,$. – 2017-01-01
  • 1
    This is the complete question..May be the question itself is wrong.. – 2017-01-01

1 Answers 1

1

HINT:

$$-(a-b)z=x(b-c)+y(c-a)$$

$$\frac{bz-cy}{b-c}=\dfrac{-b\{x(b-c)+y(c-a)\}-c(a-b)y}{(b-c)(a-b)}=?$$

  • 0
    got it..Thanks a lot for the clarification... – 2017-01-01
  • 0
    @learner, Welcome! Please follow: https://archive.org/details/higheralgebraseq00hall and https://archive.org/details/higheralgebra032813mbp – 2017-01-01
  • 0
    thanks for sharing the link. – 2017-01-01