Given that, one root of the equation \( \Large x^{2} - bx + c = 0\) is square of other root of this equation i.e., roots (\( \Large \left( \alpha , \alpha ^{2}\right) \)).
Sum of roots = \( \Large \alpha + \alpha ^{2} = - \frac{-b}{1} \)
= \( \Large \alpha \left( \alpha + 1\right) = b \) ...(i)
and product of roots = \( \Large \alpha . \alpha ^{2} = \frac{c}{1} \)
=> \( \Large \alpha ^{3} = c => \alpha = c^{1/3} \) ...(ii)
From Eqs. (i) and (ii),
\( \Large c^{1/3} \left(c^{1/3} + 1\right) = b \) ...(iii)
On ubing both sides, we get
\( \Large c \left(c^{1/3} + 1\right)^{3} = b^{3} \)
=>\( \Large c \{c + 1 + 3c^{1/3} \left(c^{1/3} + 1 \right) \} = b^{3} \)
=>\( \Large c [c + 1 + 3b] = b^{3} \) [From Eq. (iii)]
\( \Large b^{3} = 3bc + c^{2} + c \)
