(a-b)(a-b+c) = (b-a)(a-b+c)

(b-a)(c-b)(d-c) (a-b)**(b-c)(c-d)**

Can this be proven?

My opinion is that the part in bold is incorrect.

Please verify.

Thank-you!

(a-b)(a-b+c) = (b-a)(a-b+c)

(b-a)(c-b)(d-c) (a-b)**(b-c)(c-d)**

Can this be proven?

My opinion is that the part in bold is incorrect.

Please verify.

Thank-you!

Tutors, please sign in to answer this question.

It is quite simple.

Use the following fact

x - y = -1 * (y - x)

The result is as follows

(a-b)(a-b+c) -1*(b-a)(a-b+c)

----------------- = ------------------------------------

(b-a)(c-b)(d-c) [-1*(a-b)][-1*(b-c)][-1*(c-d)]

-1 * (b-a)(a-b+c) (b-a)(a-b+c)

= -------------------------- = ------------------

(-1)^{3} * (a-b)(b-c)(c-d) (a-b)(b-c)(c-d)

The last step works because -1 / (-1)^{3} = 1.

## Comments

That's pretty much it.

(-1)

^{n}= 1 if n is even and -1 if n is odd.Using this you get -1 / (-1)

^{3}= -1 / -1 = 1.