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Question Number 207664 by efronzo1 last updated on 22/May/24

$$\mathrm{If}\mathrm{p}+\mathrm{q}+\mathrm{r}=0\mathrm{and}$$$$\mathrm{A}=\frac{{({\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2)}^2}{{\left(\mathrm{pq}\right)}^2+{\left(\mathrm{qr}\right)}^2+{\left(\mathrm{rp}\right)}^2}$$$$\mathrm{B}=\frac{{\mathrm{q}}^2-\mathrm{pr}}{{\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2}.$$$$\mathrm{Find}{\mathrm{A}}^2-4\mathrm{B}$$

Answered by A5T last updated on 22/May/24

$${(p+q+r)}^2=p^2+q^2+r^2+2(pq+qr+rp)$$$$\Rightarrow p^2+q^2+r^2=-2(pq+qr+rp)$$$${\left(pq\right)}^2+{\left(qr\right)}^2+{\left(rp\right)}^2={(pq+qr+rp)}^2-2pqr(p+q+r)$$$$\Rightarrow A^2=\frac{16{(pq+qr+rp)}^4}{{(pq+qr+rp)}^4}=16$$$$q^2={(p+r)}^2=p^2+r^2+2pr\Rightarrow q^2-pr=p^2+r^2+pr$$$$p^2+q^2+r^2=2(p^2+r^2+pr)$$$$\Rightarrow B=\frac{q^2-pr}{p^2+q^2+r^2}=\frac{1}{2}$$$$\Rightarrow A-4B=16-4\left(\frac{1}{2}\right)=14$$

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