Correct Option: A

Given that, one root of the equation
x² - bx + c = 0 is square of other root of this equation i.e., roots (α, α²).
∴ Sum of roots = α + α² = -(-b)/1
⇒ α (α + 1) = b .......(i)
and product of roots= α. α² = c/1
⇒ α³ = c ⇒ = c^1/3 ....(ii)
From Eqs. (i) and (ii).
c^1/3 (c^1/3 + 1) = b ....(iii)
On cubing both sides, we get
c(c^1/3 + 1)³ = b³
⇒ c {c + 1 + 3c^1/3 (c^1/3 + 1)} = b³
⇒ c {c + 1 = 3b} = b³ [from Eq. (iii)]
⇒ b³ = 3bc + c² + c