8. Factorise each of the following:(1) 8a + b3 + 12a²b + 6ab2(i) 27 -125a – 135a + 225a2(iii) 8a3 – 63 – 12a²b + 6ab2(iv) 64a3-2763 – 144a²b + 108ab2 About the author Eden
Answer: i) 8a3 + b3 + 12a2b + 6ab2 = (2a)3 + (b)3 + 3(2a)(b)(2a + b) = (2a + b)3 ∵ a3 + b3 + 3ab(a + b) = (a + b)3 (ii) 8a3 – b3 – 12ab2 + 6ab3 = (2a)3 + (-b)3 + 3(2a)(-b)(2a – b) = (2a – b)3 ∵ a3 – b3 – 3ab(a – b) = (a – b)3 (iii) 27 – 125a3 – 135a + 225a2 OR 125a3 – 225a2 + 135a – 27 = (5a)3 + (-3)3 + 3(5a)(-3)(5a – 3) = (5a – 3)3 ∵ a3 – b3 – 3ab(a – b) = (a – b)3 (iv) 64a3 – 27b3 – 144a2b + 108ab2 = (4a)3 + (-3b)3 + 3(4a)(-3b)(4a – 3b) = (4a-3b)3 ∵ a3 – b3 – 3ab(a – b) = (a – b)3 Read more on Sarthaks.com – https://www.sarthaks.com/670350/factorise-each-of-the-following-i-8a-3-b-3-12a-2b-6ab-2?show=670353#a670353 Reply
Answer:
i) 8a3 + b3 + 12a2b + 6ab2 = (2a)3 + (b)3 + 3(2a)(b)(2a + b) = (2a + b)3 ∵ a3 + b3 + 3ab(a + b) = (a + b)3 (ii) 8a3 – b3 – 12ab2 + 6ab3 = (2a)3 + (-b)3 + 3(2a)(-b)(2a – b) = (2a – b)3 ∵ a3 – b3 – 3ab(a – b) = (a – b)3 (iii) 27 – 125a3 – 135a + 225a2 OR 125a3 – 225a2 + 135a – 27 = (5a)3 + (-3)3 + 3(5a)(-3)(5a – 3) = (5a – 3)3 ∵ a3 – b3 – 3ab(a – b) = (a – b)3 (iv) 64a3 – 27b3 – 144a2b + 108ab2 = (4a)3 + (-3b)3 + 3(4a)(-3b)(4a – 3b) = (4a-3b)3 ∵ a3 – b3 – 3ab(a – b) = (a – b)3 Read more on Sarthaks.com – https://www.sarthaks.com/670350/factorise-each-of-the-following-i-8a-3-b-3-12a-2b-6ab-2?show=670353#a670353
Answer:
(2a)³+b³+3×2a×b(2a³+b³)
a³+b³+3ab(a³+b³)=(a+b)³
=(2a+b)³