SSC CGL 18 Aug 2021 Shift 3 Paper

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Question : 63 of 100

Marks: +1, -0

If

a+b+c=0

, then what is the value of

\;{(b+c)^{2}}/{b c}+\;{(c+a)^{2}}/{c a}+\;{(a+b)^{2}}/{a b} ?

1
$-3$
$-1$
3

Solution:

Given,

a+b+c=0

If we put

a=2, b=-1, c=-1

then

2-1-1=0

It will satisfy the equation.

∴ \;{(b+c)^{2}}/{b c}+\;{(c+a)^{2}}/{c a}+\;{(a+b)^{2}}/{a b}

=\;{(-1-1)^{2}}/{-1 ×(-1)}+\;{(-1+2)^{2}}/{-1 × 2}+\;{(2-1)^{2}}/{2 ×(-1)}

=\;{(-2)^{2}}/{1}+\;{(1)^{2}}/{-2}+\;{1^{2}}/{-2}

=4-\;{1}/{2}-\;{1}/{2}=4-1=3

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