Q13 of 65 Page 137

Let us factorise the following algebraic expressions:

3a2 – b2 – c2 + 2ab – 2bc + 2ca

Given, 3a2 – b2 – c2 + 2ab – 2bc + 2ca


⇒ 4a2 – a2 – b2 – c2 + 2ab – 2bc + 2ca


⇒ (2a)2 – (a2 + b2 + c2 – 2ab + 2bc – 2ca)


⇒ (2a)2 – (a – b – c)2


[Since, from the identity III we know that, (x2 – y2) = (x – y)(x + y)]


⇒ (2a – (a – b – c))(2a + a + b + c)


⇒ (a + b + c)(3a + b + c)


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