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Question
Factorise:
a2 – 4b2 + a3 – 8b3 – (a – 2b)2
Sum
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Solution
Given, a2 – 4b2 + a3 – 8b3 – (a – 2b)2
{(a)2 – (2b)2} + {(a)3 – (2b)3} – (a – 2b)2
⇒ (a + 2b) (a – 2b) + {(a – 2b) (a2 + 2ab + 4b2)} – (a – 2b)2
Taking (a – 2b) as common from all the terms, we get,
(a – 2b) {(a + 2b) + (a2 + 2ab + 4b2) – (a – 2b)}
⇒ (a – 2b) (a + 2b + a2 + 2ab + 4b2 – a + 2b)
⇒ (a – 2b) (`\cancel(a)` + 2b + a2 + 2ab + 4b2 – `\cancel(a)` + 2b)
⇒ (a – 2b) (a2 + 2ab + 4b + 4b2)
Hence, the required is (a – 2b) (a2 + 2ab + 4b + 4b2).
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