The value of \frac{(x-y)^3+(y-z)^3+(z-x)^3}{ (x^2-y^2) ^3+ (y^2-z^2) ^3+ (z^2-x^2) ^3} where x ≠ y ≠ z, is:

Correct Answer is : 1(x+y)(y+z)(z+x)

\frac{1}{(x + y)(y + z)(z + x)}

SSC CPO 9 Nov 2022 Shift 2 Paper

Question : 124 of 200

Marks: +1, -0

The value of

\;{(x-y)^3+(y-z)^3+(z-x)^3}/{ (x^2-y^2) ^3+ (y^2-z^2) ^3+ (z^2-x^2) ^3}

where x ≠ y ≠ z, is:

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