Given (aω2+b+cω)3 + (aω+b+cω2)2 = 0 and a, b, c ∈ R. LHS = x3+y3 = (x + y) (x + ωy) (x + ω2y ) = 0 → x + y = 0 or x + ωy = 0 or x + ω2y = 0 x + y = 0 → a( ω2 + ω) + 2b + c(ω + ω2) = 0 → a(–1 ) + 2b + (–1) = 0 or a + c = 2b Similarly x + ωy = 0 → aω2 + b + cω + ω(aω+b+cω)2) = 0 → 2aω2 + b(1 + ω) + c(1 + ω) = 0 or – ω2(b + c – 2a) = 0 or b + c = 2a similarly a + b = 2c. = a, c, b are in A.P. Hence ‘A’ is correct. ax + by + c = 0 using C = 2b – a ax + by + 2b – a = 0 or a(x – 1) + b(y + 2) = 0 ∴ It passes through (1, – 2) Let f(x) = 2ax3–bx2 – cx, or (0, 1), f(0) = 0 and f(1) = 2a – (b + c) = 0 ∴ f′(x) = 0 has a root in (0, 1) ∴ 6ax2 – 2bx – c = 0 has a root in (0, 1).