# {ab(a-b)(a-c)} / {ac(b-a)(b-c)}, Simplify.

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### 4 Answers

First of all, note that you have the same letter "a" at numerator and denominator, same time, so you can simplify it. The result will be:

[b(a-b)(a-c)]/[c(b-a)(b-c)]

After that, multiply the fraction with "-1" value so that, at numerator, (a-b) will become (b-a). But, note that (b-a) paranthesys is also at denominator, so you can simplify it too. The result will be, after simplifying action:

-[b(a-c)]/c(b-c)]

`{ab(a-b)(a-c)} / {ac(b-a)(b-c)}`

They both share a on the top and bottom you should first divide by a to cancel out a, that would leave you:

`(b(a-b)(a-c)) / (c(b-a)(b-c))`

Now switch the order of b-a to -a+b,

`(b(a-b)(a-c))/ (c(-a+b)(b-c))`

Now factor out a negative to make the denominator and the nominator the same:

`(b(a-b)(a-c))/ (-c(a-b)(b-c))`

now divide by a-b to cancel them out

you should be left over with:

`(b(a-c))/(-c(b-c))`

{ab(a-b)(a-c)} / {ac(b-a)(b-c)}

First, divide this equation throughout by (a), as a is present in both side, so it turns out to be:

(b(a-b)(a-c)) / (c(b-a)(b-c))

Then, change (a-b) into (b-a), to be in the same form as the denominator Remember to put the negative sign in front of the numerator:

-(b(b-a)(a-c))/ (c(b-a)(b-c))

Divide b-a throughout top and bottom

-(b(a-c))/(c(b-c)) //

To simplify {ab(a-b)(a-c)}/{ac(b-a)(b-c)}

Observe that both numerator and denominators can be expressed as below:

Numerator factors: **a***b***(a-b)**(a-c)

Denominator factors: **a***c***(a-b)**(-1)*(b-c).

Threfore, the highest common factor(HCF) of Numerator and Denominator =**a(a-b)**.

Therfore, we can divide by the the HCF both numerator and denominator to get the simplified expresion of the given expression:

b(a-c)/{c(-1)(b-c)}, which is equivalent to

**b(a-c)/{c(c-b)}** or

**b(c-a)/{c(c-b)}**