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Question Number 61694 by naka3546 last updated on 06/Jun/19

Commented by MJS last updated on 07/Jun/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}$$

Answered by ajfour last updated on 08/Jun/19

$$p+q+r=1....\left(i\right)$$$$pqr=1....\left(ii\right)$$$$ap+bq+cr=11.....\left(iii\right)$$$$a^{2}p+b^{2}q+c^{2}r=111...\left(iv\right)$$$$p=\frac{a}{b},q=\frac{b}{c},r=\frac{c}{a}$$$$\Rightarrow a(p+\frac{q}{p}+r^2)=11$$$$a^2(p+\frac{q}{p^2}+r^3)=111$$$$\Rightarrow {(p+\frac{q}{p}+r^2)}^2=\lambda (p+\frac{q}{p^2}+r^3)$$$$\mathrm{\forall }\lambda =\frac{121}{111};Nowusing\left(ii\right)$$$${(p+\frac{q}{p}+\frac{1}{p^2{q}^2})}^2=\lambda (p+\frac{q}{p^2}+\frac{1}{p^3{q}^3})$$$$\Rightarrow \frac{{(p^3{q}^2+{pq}^3+1)}^2}{p^4{q}^4}=\lambda \frac{(p^4{q}^3+{pq}^4+1)}{p^3{q}^3}$$$$\Rightarrow {(p^3{q}^2+{pq}^3+1)}^2=\lambda pq(p^4{q}^3+{pq}^4+1)$$$$\Rightarrow {(p^3{q}^2+{pq}^3+1)}^2=\lambda p^2{q}^2(p^3{q}^2+q^3)+\lambda pq$$$$Andp+q+\frac{1}{pq}=1$$$$......$$

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