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Calculation of Intervals

x-2 y3+2 y-3 ...

Question

(x - 2y)³ + (2y - 3z)³ + (3z -x)³ = .......

A

3(x - 2y) (2y - 3z) (3z - x)

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B

(x - 2y) (2y - 3z) (3z - x)

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C

(x - 2y) (2y - 3z)

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D

2 (x - 2y) (2y - 3z) (3z - x)

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Solution

The correct option is A
3(x - 2y) (2y - 3z) (3z - x)

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Q. (x – 2y)³ + (2y – 3z)³ + (3z – x)³

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Q. Without finding the cubes, factorise

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Q. Question 38
Without finding the cubes, factorise:

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