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Question Number 52572 by Necxx last updated on 09/Jan/19
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
$${In}\:{the}\:{equation}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}.{One} \\ $$$${root}\:{is}\:{the}\:{square}\:{of}\:{the}\:{other}. \\ $$$${Show}\:{that}\:{c}\left({a}−{b}\right)^{\mathrm{3}} ={a}\left({c}−{b}\right)^{\mathrm{3}} \\ $$$$ \\ $$
Answered by math1967 last updated on 10/Jan/19
let one root=α∴ other=α^2   ∴α^2 +α=−(b/a) α^3 =(c/a)  Now ((c(a−b)^3 )/(a(c−b)^3 ))=(c/a)×((((a−b)^3 )/a^3 )/(((c−b)^3 )/a^3 ))  =(c/a)×(((1−(b/a))^3 )/(((c/a)−(b/a))^3 ))=α^3 ×(((1+α^2 +α)^3 )/((α^3 +α^2 +α)^3 )) ★  =α^3 ×(((1+α^2 +α)^3 )/(α^3 (1+α^2 +α)^3 ))=1  ∴((c(a−b)^3 )/(a(c−b)^3 ))=1 ∴c(a−b)^3 =a(c−b)^3   ★α^3 =(c/a)  α^2 +α=−(b/a)
$${let}\:{one}\:{root}=\alpha\therefore\:{other}=\alpha^{\mathrm{2}} \\ $$$$\therefore\alpha^{\mathrm{2}} +\alpha=−\frac{{b}}{{a}}\:\alpha^{\mathrm{3}} =\frac{{c}}{{a}} \\ $$$${Now}\:\frac{{c}\left({a}−{b}\right)^{\mathrm{3}} }{{a}\left({c}−{b}\right)^{\mathrm{3}} }=\frac{{c}}{{a}}×\frac{\frac{\left({a}−{b}\right)^{\mathrm{3}} }{{a}^{\mathrm{3}} }}{\frac{\left({c}−{b}\right)^{\mathrm{3}} }{{a}^{\mathrm{3}} }} \\ $$$$=\frac{{c}}{{a}}×\frac{\left(\mathrm{1}−\frac{{b}}{{a}}\right)^{\mathrm{3}} }{\left(\frac{{c}}{{a}}−\frac{{b}}{{a}}\right)^{\mathrm{3}} }=\alpha^{\mathrm{3}} ×\frac{\left(\mathrm{1}+\alpha^{\mathrm{2}} +\alpha\right)^{\mathrm{3}} }{\left(\alpha^{\mathrm{3}} +\alpha^{\mathrm{2}} +\alpha\right)^{\mathrm{3}} }\:\bigstar \\ $$$$=\alpha^{\mathrm{3}} ×\frac{\left(\mathrm{1}+\alpha^{\mathrm{2}} +\alpha\right)^{\mathrm{3}} }{\alpha^{\mathrm{3}} \left(\mathrm{1}+\alpha^{\mathrm{2}} +\alpha\right)^{\mathrm{3}} }=\mathrm{1} \\ $$$$\therefore\frac{{c}\left({a}−{b}\right)^{\mathrm{3}} }{{a}\left({c}−{b}\right)^{\mathrm{3}} }=\mathrm{1}\:\therefore{c}\left({a}−{b}\right)^{\mathrm{3}} ={a}\left({c}−{b}\right)^{\mathrm{3}} \\ $$$$\bigstar\alpha^{\mathrm{3}} =\frac{{c}}{{a}}\:\:\alpha^{\mathrm{2}} +\alpha=−\frac{{b}}{{a}} \\ $$
Commented by Necxx last updated on 10/Jan/19
thank you sir.Its clear.
$${thank}\:{you}\:{sir}.{Its}\:{clear}. \\ $$

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