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Question Number 205353 by MATHEMATICSAM last updated on 17/Mar/24

If {a x}^{2} + b x + c = 0 had two roots p and q

and p^{2} + q^{2} = p^{3} + q^{3} then show that

b^{3} - 2 a^{2} c + {a b}^{2} = 3 a b c .

Answered by sniper237 last updated on 17/Mar/24

p^{2} + q^{2} = {(p + q)}^{2} - 2 p q = \frac{b^{2}}{a^{2}} - \frac{2 c}{a}

p^{3} + q^{3} = {(p + q)}^{3} - 3 p q (p + q) = - \frac{b^{3}}{a^{3}} + \frac{3 b c}{a^{2}}

\overset{\times a^{3}}{\Rightarrow} - b^{3} + 3 a b c = {a b}^{2} - 2 a c

Answered by A5T last updated on 17/Mar/24

p + q = \frac{- b}{a}, p q = \frac{c}{a}

{(p + q)}^{2} - 2 p q = {(p + q)}^{3} - 3 p q (p + q)

\Rightarrow \frac{b^{2}}{a^{2}} - \frac{2 c}{a} = \frac{- b^{3}}{a^{3}} + \frac{3 b c}{a^{2}} \Rightarrow {a b}^{2} - 2 a^{2} c = - b^{3} + 3 a b c

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