What is \(\rm \frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}\) equal to ?
Simplifying a Complex Algebraic Fraction Expression
The problem asks us to simplify the given algebraic expression:
\(\displaystyle \frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}\)
This expression involves a sum of three algebraic fractions with variables \(x\), \(y\), and \(z\). To simplify this sum, we need to find a common denominator and combine the terms. Notice the structure of the denominators: they involve the variables \(x\), \(y\), \(z\) multiplied by differences between them.
Let's rewrite the terms to have a more consistent set of difference factors. We can use the property that \((a-b) = -(b-a)\).
- The first term is \(\displaystyle \frac{1}{x(x-y)(x-z)}\). The factors in the denominator are \(x\), \((x-y)\), and \((x-z)\).
- The second term is \(\displaystyle \frac{1}{y(y-z)(y-x)}\). We can rewrite \((y-x)\) as \(-(x-y)\). So the term becomes \(\displaystyle \frac{1}{y(y-z)(-(x-y))} = \frac{-1}{y(x-y)(y-z)}\). The factors in the denominator are \(y\), \((x-y)\), and \((y-z)\), with a negative sign.
- The third term is \(\displaystyle \frac{1}{z(z-x)(z-y)}\). We can rewrite \((z-x)\) as \(-(x-z)\) and \((z-y)\) as \(-(y-z)\). So the term becomes \(\displaystyle \frac{1}{z(-(x-z))(-(y-z))} = \frac{1}{z(x-z)(y-z)}\). The factors in the denominator are \(z\), \((x-z)\), and \((y-z)\).
So the expression can be written as:
\(\displaystyle \frac{1}{x(x-y)(x-z)} - \frac{1}{y(x-y)(y-z)} + \frac{1}{z(x-z)(y-z)}\)
To add these fractions, we find a common denominator. The common denominator should include \(x\), \(y\), \(z\), and the difference factors. A suitable common denominator is \(xyz(x-y)(y-z)(x-z)\).
Now, we rewrite each term with this common denominator:
- First term: \(\displaystyle \frac{1}{x(x-y)(x-z)} = \frac{1}{x(x-y)(x-z)} \times \frac{yz(y-z)}{yz(y-z)} = \frac{yz(y-z)}{xyz(x-y)(x-z)(y-z)}\)
- Second term: \(\displaystyle \frac{-1}{y(x-y)(y-z)} = \frac{-1}{y(x-y)(y-z)} \times \frac{xz(x-z)}{xz(x-z)} = \frac{-xz(x-z)}{xyz(x-y)(y-z)(x-z)}\)
- Third term: \(\displaystyle \frac{1}{z(x-z)(y-z)} = \frac{1}{z(x-z)(y-z)} \times \frac{xy(x-y)}{xy(x-y)} = \frac{xy(x-y)}{xyz(x-z)(y-z)(x-y)}\)
Now we can combine the numerators over the common denominator:
\(\displaystyle \frac{yz(y-z) - xz(x-z) + xy(x-y)}{xyz(x-y)(y-z)(x-z)}\)
Let's simplify the numerator: \(N = yz(y-z) - xz(x-z) + xy(x-y)\)
\(N = y^2z - yz^2 - (x^2z - xz^2) + (x^2y - xy^2)\)
\(N = y^2z - yz^2 - x^2z + xz^2 + x^2y - xy^2\)
Let's group terms by powers of \(x\):
\(N = x^2(y-z) + x(-y^2 + z^2) + (y^2z - yz^2)\)
\(N = x^2(y-z) - x(y^2 - z^2) + yz(y-z)\)
Recognize \(y^2 - z^2 = (y-z)(y+z)\):
\(N = x^2(y-z) - x(y-z)(y+z) + yz(y-z)\)
Factor out the common term \((y-z)\):
\(N = (y-z) [x^2 - x(y+z) + yz]\)
\(N = (y-z) [x^2 - xy - xz + yz]\)
Factor the quadratic expression inside the square brackets by grouping:
\(N = (y-z) [x(x-y) - z(x-y)]\)
\(N = (y-z) [(x-y)(x-z)]\)
So, the numerator simplifies to \((y-z)(x-y)(x-z)\).
Now, substitute this back into the expression:
\(\displaystyle \frac{(y-z)(x-y)(x-z)}{xyz(x-y)(y-z)(x-z)}\)
Assuming \(x, y, z\) are distinct, we can cancel the common factors \((y-z)\), \((x-y)\), and \((x-z)\) from the numerator and the denominator.
\(\displaystyle \frac{\cancel{(y-z)}\cancel{(x-y)}\cancel{(x-z)}}{xyz\cancel{(x-y)}\cancel{(y-z)}\cancel{(x-z)}} = \frac{1}{xyz}\)
The simplified expression is \(\displaystyle \frac{1}{xyz}\).
This matches one of the given options.
Revision Table: Key Steps for Algebraic Simplification
| Step | Description | Detail |
|---|---|---|
| 1 | Rewrite Terms | Adjust signs in denominators to use consistent factors like \((x-y), (y-z), (x-z)\). |
| 2 | Find Common Denominator | Identify the least common multiple of the denominators, including variables and difference factors. Here, it's \(xyz(x-y)(y-z)(x-z)\). |
| 3 | Rewrite Fractions | Multiply the numerator and denominator of each term by the factors needed to match the common denominator. |
| 4 | Combine Numerators | Add or subtract the resulting numerators over the single common denominator. |
| 5 | Simplify Numerator | Expand and factor the numerator. Look for common factors or polynomial identities. |
| 6 | Cancel Common Factors | If possible, cancel identical factors in the simplified numerator and denominator. |
Additional Information on Algebraic Fractions and Factorization
Simplifying algebraic expressions, especially those involving rational terms (fractions with polynomials), is a fundamental skill in algebra. The process often involves finding a common denominator to combine terms, similar to adding numerical fractions. Careful handling of signs in factors like \((a-b) = -(b-a)\) is crucial for correctly identifying common denominators and combining terms.
Factorization is also a key technique. In this problem, factoring the numerator \(x^2(y-z) - x(y^2 - z^2) + yz(y-z)\) allowed us to identify common factors that could be cancelled with terms in the denominator. Recognizing patterns or grouping terms strategically helps in factorization.
The expression structure in this problem is symmetric or cyclic with respect to \(x, y, z\). Problems with such cyclic symmetry often lead to elegant simplifications. The identity \(\displaystyle \frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-b)} = 0\) is a related result for just the difference terms, which shows how terms involving differences can sum to zero or a simple value.
Testing with specific numerical values (like \(x=1, y=2, z=3\)) can be a good way to check the answer, but it is not a formal proof. The algebraic simplification process provides the rigorous solution.
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Column I
Column II
a.
Equivalent fraction of \(\frac{7}{12}\) is
i.
Proper fraction
b.
Equivalent fraction of \(\frac{9}{15}\) is
ii.
Improper fraction
c.
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iii.
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d.
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iv.
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