SSC CGL 9 Aug 2015 Shift 1 Paper

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Question : 132 of 200

Marks: +1, -0

If x = 332, y = 333, z = 335, then the value of

x^3 + y^3 + z^3

– 3xyz is

10000
7000
8000
9000

Solution:

x = 332, y = 333, z = 335
∴ x + y + z = 332 + 333 + 335 = 1000
∴

x^3+ y^3+ z^3

– 3xyz =

1/2

(x + y + z) [

(x – y)^2 + (y –z)^2] + (z – x)^2

]
=

{1000}/2

[(332-333)^2

+

(333-335)^2+(335-332)^2]

= 500 (1 + 4 + 9) = 500 × 14
= 7000

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