Question Number 207664 by efronzo1 last updated on 22/May/24
$$\:\:\:\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}\:=\:\mathrm{0}\:\mathrm{and}\: \\ $$$$\:\:\:\:\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} +\left(\mathrm{rp}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\mathrm{B}=\:\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:. \\ $$$$\:\:\mathrm{Find}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}\: \\ $$
Answered by A5T last updated on 22/May/24
$$\left({p}+{q}+{r}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\left({pq}+{qr}+{rp}\right) \\ $$$$\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =−\mathrm{2}\left({pq}+{qr}+{rp}\right) \\ $$$$\left({pq}\right)^{\mathrm{2}} +\left({qr}\right)^{\mathrm{2}} +\left({rp}\right)^{\mathrm{2}} =\left({pq}+{qr}+{rp}\right)^{\mathrm{2}} −\mathrm{2}{pqr}\left({p}+{q}+{r}\right) \\ $$$$\Rightarrow{A}^{\mathrm{2}} =\frac{\mathrm{16}\left({pq}+{qr}+{rp}\right)^{\mathrm{4}} }{\left({pq}+{qr}+{rp}\right)^{\mathrm{4}} }=\mathrm{16} \\ $$$${q}^{\mathrm{2}} =\left({p}+{r}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{pr}\Rightarrow{q}^{\mathrm{2}} −{pr}={p}^{\mathrm{2}} +{r}^{\mathrm{2}} +{pr} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{2}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} +{pr}\right) \\ $$$$\Rightarrow{B}=\frac{{q}^{\mathrm{2}} −{pr}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{A}−\mathrm{4}{B}=\mathrm{16}−\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{14} \\ $$