SSC CGL 9 Aug 2015 Shift 1 Paper

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Question : 136 of 200

Marks: +1, -0

If m = – 4, n = – 2, then the value of

m^3 – 3m^2

+ 3m + 3n +

3n^2 + n^3

is

– 126
124
– 124
126

Solution:

Expression =

m^3 – 3m^2 + 3m + 3n + 3n^2 + n^3

=

m^3 – 3m^2

+ 3m – 1 +

n^3 + 3n^2

+ 3n + 1
=

(m – 1)^3 + (n + 1)^3

=

(– 4 – 1)^3 + (– 2 + 1)^3

=

(– 5)^3 + (– 1)^3

= – 125 – 1 = – 126

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