What is \(\frac{(a+b)^2}{(c-a)(c+a+b)} + \frac{(a+b)c}{c^2 + bc-a^2 - ab}\) - \(\frac{(a+2b + c)}{2(c-a)}\), \(a \neq b\), \(b \neq c\), \(c \neq a\) equal to?
Simplifying the Algebraic Expression
We are asked to simplify the following algebraic expression:
\(E = \frac{(a+b)^2}{(c-a)(c+a+b)} + \frac{(a+b)c}{c^2 + bc-a^2 - ab} - \frac{(a+2b + c)}{2(c-a)}\)
The problem provides the constraints that \(a \neq b\), \(b \neq c\), and \(c \neq a\). These constraints ensure that none of the denominators in the expression become zero.
Let's simplify this step-by-step.
Step 1: Simplifying the Denominator of the Second Term
First, we focus on the denominator of the second fraction: \(c^2 + bc - a^2 - ab\).
We can rearrange the terms to group related factors:
\(c^2 - a^2 + bc - ab\)
Now, group the terms into pairs:
\((c^2 - a^2) + (bc - ab)\)
Factor each pair. The first pair is a difference of squares, \((c^2 - a^2) = (c-a)(c+a)\). The second pair has a common factor \(b\), so \((bc - ab) = b(c-a)\).
\((c-a)(c+a) + b(c-a)\)
We can see a common factor of \((c-a)\) in both terms. Factor it out:
\((c-a)(c+a+b)\)
Interestingly, this simplified denominator is identical to the denominator of the first fraction.
Step 2: Combining the First Two Fractions
Substitute the factored denominator back into the expression:
\(E = \frac{(a+b)^2}{(c-a)(c+a+b)} + \frac{(a+b)c}{(c-a)(c+a+b)} - \frac{(a+2b + c)}{2(c-a)}\)
The first two fractions share the same denominator, \((c-a)(c+a+b)\). We can combine them by adding their numerators:
\(E = \frac{(a+b)^2 + (a+b)c}{(c-a)(c+a+b)} - \frac{(a+2b + c)}{2(c-a)}\)
Look at the numerator of the combined fraction. We can factor out the common term \((a+b)\):
\(E = \frac{(a+b)[(a+b) + c]}{(c-a)(c+a+b)} - \frac{(a+2b + c)}{2(c-a)}\)
Simplify the expression inside the brackets in the numerator:
\(E = \frac{(a+b)(a+b+c)}{(c-a)(c+a+b)} - \frac{(a+2b + c)}{2(c-a)}\)
Step 3: Simplifying the First Term Further
The term \((a+b+c)\) appears in both the numerator and the denominator of the first fraction. We can cancel this term out, provided that \(a+b+c \neq 0\). If \(a+b+c\) were 0, the original denominator \((c-a)(c+a+b)\) would be 0, which is disallowed by the problem's constraints.
After cancellation, the expression becomes:
\(E = \frac{a+b}{c-a} - \frac{a+2b + c}{2(c-a)}\)
Step 4: Finding a Common Denominator
Now we have two fractions with different denominators: \((c-a)\) and \(2(c-a)\). The least common denominator is \(2(c-a)\).
To combine these fractions, we need to make the denominator of the first fraction match the common denominator. Multiply the numerator and denominator of the first term by 2:
\(E = \frac{2(a+b)}{2(c-a)} - \frac{a+2b + c}{2(c-a)}\)
Step 5: Combining the Fractions
With the common denominator \(2(c-a)\), we can now subtract the numerators:
\(E = \frac{2(a+b) - (a+2b + c)}{2(c-a)}\)
Expand the numerator by distributing the terms:
\(E = \frac{2a + 2b - a - 2b - c}{2(c-a)}\)
Combine like terms in the numerator:
\(E = \frac{(2a - a) + (2b - 2b) - c}{2(c-a)}\)
\(E = \frac{a - c}{2(c-a)}\)
Step 6: Final Simplification
Observe the numerator \(a-c\) and the denominator term \(c-a\). They are negatives of each other. We can rewrite the numerator as \(-(c-a)\):
\(E = \frac{-(c-a)}{2(c-a)}\)
The problem states that \(c \neq a\), meaning \((c-a)\) is not zero. Therefore, we can cancel the \((c-a)\) term from the numerator and the denominator.
\(E = -\frac{1}{2}\)
Conclusion
After simplifying the given algebraic expression step-by-step, we find that it equals \(-\frac{1}{2}\).
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