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Question Number 52572 by Necxx last updated on 09/Jan/19

I n t h e e q u a t i o n {a x}^{2} + b x + c = 0 . O n e

r o o t i s t h e s q u a r e o f t h e o t h e r .

S h o w t h a t c {(a - b)}^{3} = a {(c - b)}^{3}

Answered by math1967 last updated on 10/Jan/19

l e t o n e r o o t = α ∴ o t h e r = α^{2}

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

N o w \frac{c {(a - b)}^{3}}{a {(c - b)}^{3}} = \frac{c}{a} \times \frac{\frac{{(a - b)}^{3}}{a^{3}}}{\frac{{(c - b)}^{3}}{a^{3}}}

= \frac{c}{a} \times \frac{{(1 - \frac{b}{a})}^{3}}{{(\frac{c}{a} - \frac{b}{a})}^{3}} = α^{3} \times \frac{{(1 + α^{2} + α)}^{3}}{{(α^{3} + α^{2} + α)}^{3}} ★

= α^{3} \times \frac{{(1 + α^{2} + α)}^{3}}{α^{3} {(1 + α^{2} + α)}^{3}} = 1

∴ \frac{c {(a - b)}^{3}}{a {(c - b)}^{3}} = 1 ∴ c {(a - b)}^{3} = a {(c - b)}^{3}

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

Commented by Necxx last updated on 10/Jan/19

t h a n k y o u s i r . I t s c l e a r .

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