SSC CPO SI and ASI 2014 Paper

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Question : 135 of 200

Marks: +1, -0

The numerical value of

{(a-b)^{2}}/{(b-c)(c-a)}+ {(b-c)^{2}}/{(c-a)(a-c)}+ {(c-a)^{2}}/{(a-b)(b-c)}

is

[a≠ b ≠ c]

0
1
$1/ 3$
3

Solution:

(4) Let x = a – b, y = b – c and z = c – a. Since, x + y + z = a – b + b – c + c – a = 0, therefore x

^{3}

+ y

^{3}

+ z

^{3}

= 3xyz
Therefore,

{(a-b)^{2}}/{(b-c)(c-a)}+ {(b-c)^{2}}/{(c-a)(a-c)}+ {(c-a)^{2}}/{(a-b)(b-c)}

= {(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}/{(a-b)(b-c)(c-a)}

= {3(a-b)(b-c)(c-a)}/{(a-b)(b-c)(c-a)}=3

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