Question Number 153161 by mathdanisur last updated on 05/Sep/21
![If P(x)=aX^3 +bX+c ∈ Q[X] ; (a≠0) has the roots x_1 , x_2 and x_3 such that x_1 = x_2 x_3 then prove that x_1 = 0 if and only if b=c](Q153161.png)
$$\mathrm{If}\:\:\mathrm{P}\left(\mathrm{x}\right)=\mathrm{aX}^{\mathrm{3}} +\mathrm{bX}+{c}\:\in\:\mathrm{Q}\left[\mathrm{X}\right]\:\:;\:\:\left(\mathrm{a}\neq\mathrm{0}\right) \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\:\:\mathrm{x}_{\mathrm{1}} \:,\:\mathrm{x}_{\mathrm{2}} \:\:\mathrm{and}\:\:\mathrm{x}_{\mathrm{3}} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}_{\mathrm{1}} \:=\:\mathrm{x}_{\mathrm{2}} \mathrm{x}_{\mathrm{3}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{x}_{\mathrm{1}} \:=\:\mathrm{0}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{b}=\mathrm{c} \\ $$
Answered by mr W last updated on 05/Sep/21
$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} \left({x}_{\mathrm{2}} +{x}_{\mathrm{3}} \right)+{x}_{\mathrm{2}} {x}_{\mathrm{3}} =\frac{{b}}{{a}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} =−\frac{{c}}{{a}} \\ $$$$ \\ $$$${with}\:{x}_{\mathrm{1}} ={x}_{\mathrm{2}} {x}_{\mathrm{3}} : \\ $$$$−{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{1}} =\frac{{b}}{{a}} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} =−\frac{{c}}{{a}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\frac{{b}−{c}}{{a}} \\ $$$${x}_{\mathrm{1}} =\mathrm{0}\:{if}\:{and}\:{only}\:{if}\:{b}−{x}=\mathrm{0}\:\Rightarrow{b}={c} \\ $$
Commented by mathdanisur last updated on 05/Sep/21
$$\mathrm{veri}\:\mathrm{nice}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{ser} \\ $$