Difficulty: Easy
Correct Answer: 3
Explanation:
Introduction / Context:This question tests basic exponent evaluation and cube roots. The right-hand side is a simple difference of squares, and once it is computed, the equation becomes x^3 = 27, whose cube root is a standard value. The key is to compute 6^2 and 3^2 correctly and not confuse x^3 with 3x.
Given Data / Assumptions:
Concept / Approach:Evaluate powers: 6^2 and 3^2. Subtract to get a number. Then solve x^3 = that number by taking the cube root (the number whose cube equals the given value).
Step-by-Step Solution:
Step 1: Compute 6^2 = 6*6 = 36. Step 2: Compute 3^2 = 3*3 = 9. Step 3: Subtract: 6^2 - 3^2 = 36 - 9 = 27. Step 4: So the equation is x^3 = 27. Step 5: The cube root of 27 is 3 because 3^3 = 3*3*3 = 27. Step 6: Therefore x = 3.Verification / Alternative check:Plug back: If x = 3, then x^3 = 27. RHS is 36 - 9 = 27. Both sides match exactly, so x = 3 is correct.
Why Other Options Are Wrong:
6: 6^3 = 216, not 27. 9: 9^3 = 729, not 27. 1: 1^3 = 1, not 27. -3: (-3)^3 = -27, wrong sign.Common Pitfalls:Misreading x^3 as 3x, computing 6^2 - 3^2 as 3^2 (wrong), or forgetting cube root vs square root.
Final Answer:3
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