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Sum of cubes of three cyclic differences (x - y), (y - z), (z - x)

This type of question, asked 2 times: 13 Sep 2024 S1, 17 Sep 2025 S3. All 2 questions below, oldest first.

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  1. Question 1

    13 Sep 2024 S1

    56
    Simplify the following expression. (x−y)3+(y−z)3+(z−x)3(x−y)(y−z)(z−x)\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(x - y)(y - z)(z - x)}

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  2. Question 2

    17 Sep 2025 S3

    56
    Given x+y+z=0x + y + z = 0, and x3+y3+z3=3xyzx^3 + y^3 + z^3 = 3xyz, evaluate (x−y)3+(y−z)3+(z−x)3(x - y)^3 + (y - z)^3 + (z - x)^3.

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