Factorise using identities 8a3 + b3 - 6ab + 1 Share with your friends Share 0 Abhishek Kumar answered this 0 View Full Answer Neha Thomas answered this 8a3+ b3 +1 - 6ab = (2a)3 +b3 + 1×1×1 - 3 ×2a × b×1 = (2a+b+1) {(2b)2+(b)2+1×1-2a×b-b×1 - 1×2a} = (2a+b+1) (4b2+ b2 + 1 -2ab-b-2a) 1 Neha Thomas answered this 8a³ + b³ + 1 - 6ab = (2a)³ + (b)³ + 1³ - 3(2a)(b)(1) = [2a + b + 1][(2a)²+ b² + 1² - (2a)b - (2a)1 - b(1)] = [2a + b + 1][(4a²+ b² + 1 - 2ab - 2a - b] identity used = x³ + y³ + z³– 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx) 2
