Q. 53.9( 12 Votes )

# Let us factorise the following algebraic expressions:

(3a – 2b)^{3} + (2b – 5c)^{3} + (5c – 3a)^{3}

Answer :

Let us consider (3a - 2b) = x, (2b - 5c) = y, (5c - 3a) = z

So x + y + z = 3a – 2b + 2b – 5c + 5c – 3a = 0

(3a – 2b) ^{3} + (2b – 5c) ^{3} + (5c – 3a) ^{3} = x^{3} + y^{3} + z^{3}

Since x + y + z = 0, hence

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

⇒ (3a – 2b) ^{3} + (2b – 5c) ^{3} + (5c – 3a) ^{3} = 3×(3a – 2b)×(2b – 5c)× (5c – 3a)

3(3a – 2b) (2b – 5c) (5c – 3a)

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