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