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Cube identity on (x-y), (y-z), (z-x), whose sum is zero

This type of question, asked 4 times: CDS 2 2017, CDS 1 2019, CDS 2 2019, CDS 1 2025. All 4 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    CDS 2 2017

    34
    What is (x−y)(y−z)(z−x)(x−y)3+(y−z)3+(z−x)3\frac{(x - y) (y - z) (z - x)}{(x - y)^3 + (y - z)^3 + (z - x)^3} equal to ?

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  2. Question 2

    CDS 1 2019

    98
    What is the value of (x−y)3+(y−z)3+(z−x)39(x−y)(y−z)(z−x)\frac{(x-y)^3 + (y-z)^3 + (z-x)^3}{9(x-y)(y-z)(z-x)} ?

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  3. Question 3

    CDS 2 2019

    46
    What is (x−y)3+(y−z)3+(z−x)33(x−y)(y−z)(z−x)\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{3(x - y)(y - z)(z - x)} equal to ?

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  4. Question 4

    CDS 1 2025

    36
    What is (a−b)2(b−c)(c−a)+(b−c)2(c−a)(a−b)+(c−a)2(a−b)(b−c)−3\frac{(a-b)^2}{(b-c)(c-a)} + \frac{(b-c)^2}{(c-a)(a-b)} + \frac{(c-a)^2}{(a-b)(b-c)} - 3 equal to, where a≠b≠ca \ne b \ne c ?

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