Certainly! Let's solve the expression step by step.
The given expression is:
[tex]\[
\frac{a}{(a-c)(a-b)} + \frac{b}{(b-c)(b-a)}
\][/tex]
To simplify this, we'll start by finding a common denominator. Notice that the denominators of these two fractions are [tex]\((a-c)(a-b)\)[/tex] and [tex]\((b-c)(b-a)\)[/tex]. These look very similar but with a slight difference. We have:
[tex]\[
(a - c)(a - b) \quad \text{and} \quad (b - c)(b - a)
\][/tex]
To proceed, it's important to recognize the relationships between these terms. First, observe that [tex]\((b-a)\)[/tex] is simply [tex]\(-(a-b)\)[/tex]. Therefore, [tex]\((b-c)(b-a)\)[/tex] can be rewritten using the negative sign:
[tex]\[
(b - c)(b - a) = (b - c)(-(a - b)) = -(b - c)(a - b)
\][/tex]
Thus, the original expression can be rewritten as:
[tex]\[
\frac{a}{(a - c)(a - b)} - \frac{b}{(b - c)(a - b)}
\][/tex]
Now we have a common denominator for both fractions, which is [tex]\((a - b)(a - c)(b - c)\)[/tex]. We can combine these two fractions:
[tex]\[
\frac{a(b - c) - b(a - c)}{(a - c)(a - b)(b - c)}
\][/tex]
Simplify the numerator:
[tex]\[
a(b - c) - b(a - c) = ab - ac - ba + bc = bc - ac
\][/tex]
So, the expression now is:
[tex]\[
\frac{bc - ac}{(a - c)(a - b)(b - c)}
\][/tex]
Factor out [tex]\(c\)[/tex] from the numerator:
[tex]\[
\frac{c(b - a)}{(a - c)(a - b)(b - c)}
\][/tex]
Notice that [tex]\(b - a\)[/tex] is [tex]\( -(a - b) \)[/tex]:
[tex]\[
\frac{c(- (a - b))}{(a - c)(a - b)(b - c)} = \frac{-c(a - b)}{(a - c)(a - b)(b - c)}
\][/tex]
Cancel [tex]\((a - b)\)[/tex] in the numerator and the denominator:
[tex]\[
\frac{-c}{(a - c)(b - c)}
\][/tex]
Thus, the simplified form of the expression is:
[tex]\[
\frac{-c}{(a - c)(b - c)}
\][/tex]
Rewriting it in a more standard form, we have:
[tex]\[
-\frac{c}{(a - c)(b - c)}
\][/tex]
Therefore, the simplified form of the expression [tex]\(\frac{a}{(a-c)(a-b)} + \frac{b}{(b-c)(b-a)}\)[/tex] is:
[tex]\[
-\frac{c}{(ab - ac - b*c + c^2)}
\][/tex]
