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SSC CHSL 2021 PYQ — Question 66

SSC CHSL - 6 Aug 2021 - Shift 1 • Quantitative Aptitude • Q66

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If a + b - c = 0, then what is the value of (b−c)24bc+(c−a)24ca+(a+b)24ab\frac{{{{\left( {b - c} \right)}^2}}}{{4bc}}+\frac{{{{\left( {c - a} \right)}^2}}}{{4ca}}+\frac{{{{\left( {a + b} \right)}^2}}}{{4ab}} ?

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Correct answer: D

Explanation

The correct answer is option 4 i.e. - 3/4.

If a + b + (-c) = 0

Then,

⇒ a3 + b3 - c3 = - 3abc

Dividing both sides by 4abc

⇒ a2/4bc + b2/4ac - c2/4ab = -3

⇒ a2/bc + b2/ac - c2/ab = -3/4

⇒ (b−c)24bc+(c−a)24ca+(a+b)24ab=−34\frac{{{{\left( {b - c} \right)}^2}}}{{4bc}}+\frac{{{{\left( {c - a} \right)}^2}}}{{4ca}}+\frac{{{{\left( {a + b} \right)}^2}}}{{4ab}} = -\dfrac{3}{4}