RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

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Question : 31 of 120

Marks: +1, -0

I f

(a – b) = 3, (b – c) = 5

and

(c – a) = 1

, then the value of

{a^3+b^3+c^3-3abc}/{a+b+c}

is

17.5
20.5
10.5
15.5

Solution:

(a – b) = 3, (b – c) = 5, (c – a) = 1

a^3 + b^3 + c^3 – 3abc

= (a + b + c)[a^2 + b^2 + c^2 – ab – bc – ca]

= 1/2(a+ b +c) [2a^2 +2b^2 +2c^2 -2ab -2bc -2ca]

= 1/2(a+b+ c )[(a -b)^2 +(b -c)^2 +(c -a)^2 ]

∴

{a^3 +b^3+ c^3 -3abc}/{a +b+ c}

= 1/2 [(a-b)^2+(b-c)^2+(c-a)^2]

=1/2[9+25+1]

= 17.5

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