Understanding the Algebraic Expression

The question asks for the value of a given algebraic expression which is a sum of three rational terms. Each term involves squares of variables or expressions, suggesting the use of algebraic identities, particularly the difference of squares formula.

The expression is:

\(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2}+\frac{q^2-(p-r)^2}{(p+q)^2-r^2}+\frac{r^2-(p-q)^2}{(q+r)^2-p^2}\)

Key Concept: Difference of Squares

We will use the algebraic identity for the difference of two squares:

\(a^2 - b^2 = (a+b)(a-b)\)

This identity is fundamental for simplifying the numerator and the denominator of each term in the expression.

Simplifying the First Term

The first term is \(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2}\).

Let's simplify the numerator:

Numerator: \(p^2 - (q-r)^2\)

Using the difference of squares with \(a=p\) and \(b=(q-r)\):

\(p^2 - (q-r)^2 = (p + (q-r))(p - (q-r))\)

\(= (p+q-r)(p-q+r)\)

Now, let's simplify the denominator:

Denominator: \((p+r)^2 - q^2\)

Using the difference of squares with \(a=(p+r)\) and \(b=q\):

\((p+r)^2 - q^2 = ((p+r) + q)((p+r) - q)\)

\(= (p+r+q)(p+r-q)\)

\(= (p+q+r)(p-q+r)\)

So the first term becomes:

\(\frac{(p+q-r)(p-q+r)}{(p+q+r)(p-q+r)}\)

Assuming \((p-q+r) \ne 0\), we can cancel the common factor \((p-q+r)\) from the numerator and the denominator.

First Term Simplified = \(\frac{p+q-r}{p+q+r}\)

Simplifying the Second Term

The second term is \(\rm\frac{q^2-(p-r)^2}{(p+q)^2-r^2}\).

Numerator: \(q^2 - (p-r)^2\)

Using difference of squares with \(a=q\) and \(b=(p-r)\):

\(q^2 - (p-r)^2 = (q + (p-r))(q - (p-r))\)

\(= (q+p-r)(q-p+r)\)

\(= (p+q-r)(-p+q+r)\)

Denominator: \((p+q)^2 - r^2\)

Using difference of squares with \(a=(p+q)\) and \(b=r\):

\((p+q)^2 - r^2 = ((p+q) + r)((p+q) - r)\)

\(= (p+q+r)(p+q-r)\)

So the second term becomes:

\(\frac{(p+q-r)(-p+q+r)}{(p+q+r)(p+q-r)}\)

Assuming \((p+q-r) \ne 0\), we can cancel the common factor \((p+q-r)\) from the numerator and the denominator.

Second Term Simplified = \(\frac{-p+q+r}{p+q+r}\)

Simplifying the Third Term

The third term is \(\rm\frac{r^2-(p-q)^2}{(q+r)^2-p^2}\).

Numerator: \(r^2 - (p-q)^2\)

Using difference of squares with \(a=r\) and \(b=(p-q)\):

\(r^2 - (p-q)^2 = (r + (p-q))(r - (p-q))\)

\(= (r+p-q)(r-p+q)\)

\(= (p-q+r)(-p+q+r)\)

Denominator: \((q+r)^2 - p^2\)

Using difference of squares with \(a=(q+r)\) and \(b=p\):

\((q+r)^2 - p^2 = ((q+r) + p)((q+r) - p)\)

\(= (q+r+p)(q+r-p)\)

\(= (p+q+r)(-p+q+r)\)

So the third term becomes:

\(\frac{(p-q+r)(-p+q+r)}{(p+q+r)(-p+q+r)}\)

Assuming \((-p+q+r) \ne 0\), we can cancel the common factor \((-p+q+r)\) from the numerator and the denominator.

Third Term Simplified = \(\frac{p-q+r}{p+q+r}\)

Combining the Simplified Terms

Now we add the three simplified terms:

Value = \(\frac{p+q-r}{p+q+r} + \frac{-p+q+r}{p+q+r} + \frac{p-q+r}{p+q+r}\)

Since all terms have the same denominator \((p+q+r)\), we can add the numerators directly:

Value = \(\frac{(p+q-r) + (-p+q+r) + (p-q+r)}{p+q+r}\)

Combine the terms in the numerator:

Numerator = \(p + q - r - p + q + r + p - q + r\)

Group like terms:

Numerator = \((p - p + p) + (q + q - q) + (-r + r + r)\)

Numerator = \(p + q + r\)

Final Result

Substituting the simplified numerator back into the expression:

Value = \(\frac{p+q+r}{p+q+r}\)

Assuming \((p+q+r) \ne 0\), the value of the expression is:

Value = \(1\)

Summary of Terms

Here is a summary of how each term simplifies:

• First term: \(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2} \rightarrow \frac{(p+q-r)(p-q+r)}{(p+q+r)(p-q+r)} = \frac{p+q-r}{p+q+r}\) (if \(p-q+r \ne 0\))
• Second term: \(\rm\frac{q^2-(p-r)^2}{(p+q)^2-r^2} \rightarrow \frac{(p+q-r)(-p+q+r)}{(p+q+r)(p+q-r)} = \frac{-p+q+r}{p+q+r}\) (if \(p+q-r \ne 0\))
• Third term: \(\rm\frac{r^2-(p-q)^2}{(q+r)^2-p^2} \rightarrow \frac{(p-q+r)(-p+q+r)}{(p+q+r)(-p+q+r)} = \frac{p-q+r}{p+q+r}\) (if \(-p+q+r \ne 0\))

Sum = \(\frac{p+q-r + (-p+q+r) + (p-q+r)}{p+q+r} = \frac{p+q+r}{p+q+r} = 1\) (if \(p+q+r \ne 0\))

Revision Table: Useful Algebraic Identities

Additional Information: Conditions for Validity

The simplification relies on cancelling common factors in the numerators and denominators of the fractions. This cancellation is valid only when the factors being cancelled are not equal to zero. For example, in the first term, we cancelled \((p-q+r)\), so the simplification is valid if \(p-q+r \ne 0\). Similarly, for the second term, we assume \(p+q-r \ne 0\), and for the third term, we assume \(-p+q+r \ne 0\).

Furthermore, the final step involves dividing by the common denominator \((p+q+r)\). This operation is only defined if the denominator is not zero, i.e., \(p+q+r \ne 0\). If \(p+q+r = 0\), the original expression involves division by zero in the denominators of the simplified terms, and thus the expression would be undefined.