SSC CGL Tier II 21 Sep 2014 Paper

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Question : 18 of 100

Marks: +1, -0

If ab + bc + ca = 0 then the value of

( {1}/{a^{2}-b c}+ {1}/{b^{2}-c a}+ {1}/{c^{2}-a b} )

is

0
1
3
a + b + c

Solution:

(1)

ab + bc + ca = 0

⇒ ab + ca = – bc

∴ a^{2}-b c=a^{2}+a b+a c

=a(a+b+c)

Similarly.

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

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

∴

{1}/{a^{2}-b c}+ {1}/{b^{2}-c a}+ {1}/{c^{2}-a b}

= {1}/{a(a+b+c)}+ {1}/{b(a+b+c)}+ {1}/{c(a+b+c)}

= {1}/{(a+b+c)} ( {1}/{a}+ {1}/{b}+ {1}/{c} )

= {1}/{a+b+c} ( {b c+c a+a b}/{a b c} )

= {1}/{a+b+c} ×{0}/{a b c}=0

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