If x and y are real numbers such that x + y = 5 and x^3 + y^3 = 35, what is the positive value of the difference between x and y, that is, |x − y|?

Introduction / Context:
This algebra question uses symmetric expressions in two variables. We are given the sum x + y and the sum of cubes x^3 + y^3, and we are asked to find the absolute difference |x − y|. Rather than solving for x and y individually, we use identities that relate sums and products of numbers to their powers. Such problems are popular in aptitude exams because they reward knowledge of algebraic identities and efficient manipulation of expressions.

Given Data / Assumptions:

• x and y are real numbers.
• x + y = 5.
• x^3 + y^3 = 35.
• We must find |x − y|, the positive difference between x and y.
• No additional information about ordering of x and y is required because we take the absolute value.

Concept / Approach:
We use the identity for the sum of cubes: x^3 + y^3 = (x + y)^3 − 3xy(x + y). The given values allow us to set up an equation involving xy. Once we know xy, we can use another identity that connects x − y with x + y and xy, namely (x − y)^2 = (x + y)^2 − 4xy. Taking the square root then yields |x − y|. This avoids solving a quadratic for x or y and keeps the work clean and quick.

Step-by-Step Solution:
Step 1: Start from the identity x^3 + y^3 = (x + y)^3 − 3xy(x + y).Step 2: Substitute x + y = 5 and x^3 + y^3 = 35.Step 3: The equation becomes 35 = 5^3 − 3xy × 5.Step 4: Compute 5^3 = 125, so 35 = 125 − 15xy.Step 5: Rearrange to solve for xy: 15xy = 125 − 35 = 90.Step 6: Therefore xy = 90/15 = 6.Step 7: Use the identity (x − y)^2 = (x + y)^2 − 4xy.Step 8: Substitute x + y = 5 and xy = 6: (x − y)^2 = 5^2 − 4 × 6 = 25 − 24 = 1.Step 9: Taking square roots, |x − y| = √1 = 1.

Verification / Alternative check:
We can find one explicit pair (x, y) that satisfies the conditions. Since x + y = 5 and xy = 6, the numbers are roots of t^2 − 5t + 6 = 0, which factors as (t − 2)(t − 3) = 0. So x and y must be 2 and 3 in some order. With x = 3 and y = 2, x^3 + y^3 = 27 + 8 = 35, which matches the given condition. The difference |x − y| = |3 − 2| = 1, confirming our result.

Why Other Options Are Wrong:
The option 0 would require x = y, but that would give x + y = 10 or x^3 + y^3 inconsistent with the data. The options 5 and 6 represent much larger differences and do not fit the derived relation (x − y)^2 = 1. The option 2 would correspond to a squared difference of 4, again inconsistent with the computation. Only 1 matches the algebraic relations obtained from the identities and the numeric verification.

Common Pitfalls:
Some learners try to guess x and y directly without using identities, which may work in simple cases but is not a reliable general strategy. Others misuse the formula for x^3 + y^3 or incorrectly expand (x + y)^3, leading to algebraic errors. Forgetting to use the absolute value and reporting x − y instead of |x − y| is another minor mistake. Sticking carefully to the identities and checking each step avoids these issues.

Final Answer:
The positive difference between x and y is |x − y| = 1.