If l + m + n = 9 and l^2 + m^2 + n^2 = 31 for real numbers l, m and n, find the value of the pairwise product sum lm + mn + nl.

Introduction / Context:
This algebra question tests understanding of symmetric expressions and the identity that links the square of a sum to the sum of squares and pairwise products. Problems like this appear frequently in aptitude tests to check comfort with algebraic manipulation without directly solving for individual variables. The goal is to find lm + mn + nl using the given values of l + m + n and l^2 + m^2 + n^2.

Given Data / Assumptions:

• l, m and n are real numbers.

• l + m + n = 9.

• l^2 + m^2 + n^2 = 31.

• We need lm + mn + nl.

Concept / Approach:
The key identity is (l + m + n)^2 = l^2 + m^2 + n^2 + 2(lm + mn + nl). This identity expresses the square of a sum as the sum of squares plus twice the sum of pairwise products. We already know both the left side and the sum of squares on the right side, so we can directly solve for lm + mn + nl. This avoids any need to find the individual values of l, m and n.

Step-by-Step Solution:
Step 1: Write the identity: (l + m + n)^2 = l^2 + m^2 + n^2 + 2(lm + mn + nl). Step 2: Substitute the given values: (9)^2 = 31 + 2(lm + mn + nl). Step 3: Compute 9^2 = 81. Step 4: Set up the equation: 81 = 31 + 2(lm + mn + nl). Step 5: Subtract 31 from both sides: 81 − 31 = 2(lm + mn + nl). Step 6: Compute 81 − 31 = 50, so 50 = 2(lm + mn + nl). Step 7: Divide by 2: lm + mn + nl = 50 / 2 = 25.

Verification / Alternative check:
We can check the identity by expanding (l + m + n)^2 directly: (l + m + n)^2 = l^2 + m^2 + n^2 + 2lm + 2mn + 2nl. This matches the identity used. Since the algebraic manipulation is straightforward and depends only on substitution and rearranging, the result lm + mn + nl = 25 is reliable. There is no need to solve for l, m and n individually, which might lead to unnecessary complexity or multiple possible triples, while the symmetric sum remains fixed.

Why Other Options Are Wrong:
Option 22 and 31 are values that do not satisfy the equation 81 = 31 + 2S, where S represents lm + mn + nl. Option 50 incorrectly treats 50 as the final sum without dividing by 2. Option −25 would require a negative symmetric sum, yet 81 = 31 + 2(−25) would give only 81 = −19, which is clearly false. Therefore, only 25 is consistent with the identity and the given numerical values.

Common Pitfalls:
Students sometimes forget the factor 2 in the identity and write (l + m + n)^2 = l^2 + m^2 + n^2 + (lm + mn + nl), which leads to the incorrect answer 50. Another common error is substituting incorrectly, for example squaring each term individually without including cross terms. Remembering the full expansion and carefully isolating the required sum helps avoid these mistakes. Practising similar problems reinforces the use of this powerful identity in algebra.

Final Answer:
The value of the pairwise product sum lm + mn + nl is 25.