Question

[((a ^ 2 - 25b ^ 2) ^ 3 + (25b ^ 2 - 9c ^ 2) ^ 3 + (9c ^ 2 - a ^ 2) ^ 3)/((a - 5b) ^ 3 + (5b - 3c) ^ 3 + (3c - a) ^ 3)]​

Answer 1

The factorization is  [(a + 5b)/(a - 5b)]³ + [(5b + 3c)(5b - 3c)]³ + [(3c + a)(3c - a)]³

To answer the question, w e need to know what factorization is

What is factorization?

Factorization is the breaking down of a larger expression into smaller expression by grouping it into its factors.

So, [{(a² - 25b²}³ + {25b² - 9c²}³ + {9c² - a²}³)/{(a - 5b)³ + (5b - 3c)³ + (3c - a)³}]​

= [{(a² - (5b)²}³ + {(5b)² - (3c)²}³ + {(3c)² - a²)})/{(a - 5b)³ + (5b - 3c)³ + (3c - a)³}]​

Using the fact that x² - y² = (x + y)(x - y) in the terms in the numerator, we have

[{(a - 5b)(a + 5b)}³ + {(5b - 3c)(5b + 3c)}³ + {(3c - a)(3c + a)}³/{(a - 5b)³ + (5b - 3c)³ + (3c - a)³}]​

Factorizing out the denominator, we have

= [{(a + 5b)/(a - 5b)}³ + {(5b + 3c)/(5b - 3c)}³ + {(3c + a)/(3c - a)}³[(a - 5b)³ + (5b - 3c)³ + (3c - a)³]/{(a - 5b)³ + (5b - 3c)³ + (3c - a)³}]​

Cancelling out the denominator, we have

= [(a + 5b)/(a - 5b)]³ + [(5b + 3c)(5b - 3c)]³ + [(3c + a)(3c - a)]³

So, the factorization is  [(a + 5b)/(a - 5b)]³ + [(5b + 3c)(5b - 3c)]³ + [(3c + a)(3c - a)]³

Learn more about factorization here:

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