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SSC CHSL 2020 PYQ — Question 68

SSC CHSL - 19th October 2020 - Shift 1 • Quantitative Aptitude • Q68

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The value of [(a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3] ÷ [(a – b)3 + (b – c)3 + (c – a)3 is equal to:

(Given a ≠ b ≠ c)

Show answer & explanation

Correct answer: A

Explanation

The correct answer is Option 1 i.e (a + b) (b + c) (c + a).

Understanding/Application

if a + b + c = 0 ; a³ + b³ + c³ = 3abc
(a² - b²) + (b² - c²) + (c² - a²) = 0
(a² - b²)³ + (b² - c²)³ + (c² - a²)³ = 3(a² - b²)(b² - c²)(c² - a²)

Similarly, 
(a - b) + (b - c) + (c - a) = 0
(a - b)³ + (b - c)³ + (c - a)³ = 3(a - b)(b - c)(c - a)

Now,

[(a² - b²)³ + (b² - c²)³ + (c² - a²)³ ]/[(a - b)³ + (b - c)³ + (c - a)³]
= 3(a² - b²)(b² - c²)(c² - a²)/3(a - b)(b - c)(c - a)

= (a + b)(b + c)(c + a)