Simplify: {(a²-b²)³+(b²-c²)³+(c²-a²)³}{(a-b)³+(b-c)³+(c-a)³}

Correct Answer is : (a+b)(b+c)(c+a)(a+b)(b+c)(c+a)(a+b)(b+c)(c+a)

Correct Answer is : (a+b)(b+c)(c+a)(a+b)(b+c)(c+a)(a+b)(b+c)(c+a)

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SSC CHSL 14 Nov 2025 Shift 3 Paper

Question : 74 of 100

Marks: +1, -0

Simplify:

\frac{(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}

Solution:

Concept:

x + y + z = 0

, then

x^{3} + y^{3} + z^{3} = 3xyz

Explanation:

First, consider the numerator:

(a^{2} - b^{2}) + (b^{2} - c^{2}) + (c^{2} - a^{2}) = 0

Hence, numerator

= 3(a^{2} - b^{2})(b^{2} - c^{2})(c^{2} - a^{2})

Similarly, denominator:

(a - b) + (b - c) + (c - a) = 0

Hence, denominator

= 3(a - b)(b - c)(c - a)

Now the expression becomes

\frac{3(a^{2} - b^{2})(b^{2} - c^{2})(c^{2} - a^{2})}{3(a - b)(b - c)(c - a)}

Factor each difference of squares:

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

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

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

Cancel the common factor

3(a-b)(b-c)(c-a)

from numerator and denominator.

The result is

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

Answer:

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

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