In the equation ax^2 +bx+c=0.One root is the square of the other. Show that c(a−b)^3 =a(c−b)^3


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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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