Question Number 116221 by bobhans last updated on 02/Oct/20
$$\mathrm{Given}\:\alpha,\beta\:\mathrm{and}\:\varphi\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{px}^{\mathrm{2}} +\mathrm{qx}−\mathrm{pq}\:=\:\mathrm{0}\:. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+\frac{\beta}{\varphi}+\frac{\varphi}{\beta}+\frac{\alpha}{\varphi}+\frac{\varphi}{\alpha}=? \\ $$
Answered by TANMAY PANACEA last updated on 02/Oct/20
$$\left(\frac{\beta}{\alpha}+\frac{\varphi}{\alpha}+\mathrm{1}\right)+\left(\frac{\alpha}{\beta}+\frac{\varphi}{\beta}+\mathrm{1}\right)+\left(\frac{\alpha}{\varphi}+\frac{\beta}{\varphi}+\mathrm{1}\right)−\mathrm{3} \\ $$$$=\left(\alpha+\beta+\varphi\right)\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\varphi}\right)−\mathrm{3} \\ $$$$=\left(\alpha+\beta+\varphi\right)\left(\frac{\alpha\beta+\beta\varphi+\alpha\varphi}{\alpha\beta\varphi}\right)−\mathrm{3} \\ $$$$={p}\left(\frac{{q}}{{pq}}\right)−\mathrm{3}=−\mathrm{2} \\ $$
Answered by ruwedkabeh last updated on 02/Oct/20
$$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+\frac{\beta}{\varphi}+\frac{\varphi}{\beta}+\frac{\alpha}{\varphi}+\frac{\varphi}{\alpha} \\ $$$$=\frac{\alpha^{\mathrm{2}} \varphi+\beta^{\mathrm{2}} \varphi+\alpha\beta^{\mathrm{2}} +\alpha\varphi^{\mathrm{2}} +\alpha^{\mathrm{2}} \beta+\beta\varphi^{\mathrm{2}} }{\alpha\beta\varphi} \\ $$$$=\frac{\left(\alpha+\beta+\varphi\right)\left(\alpha\beta+\alpha\varphi+\beta\varphi\right)−\mathrm{3}\alpha\beta\varphi}{\alpha\beta\varphi} \\ $$$$=\frac{\left({p}\right)\left({q}\right)−\mathrm{3}{pq}}{{pq}} \\ $$$$=−\mathrm{2} \\ $$$$ \\ $$
Answered by bobhans last updated on 02/Oct/20
