Question Number 211330 by RojaTaniya last updated on 06/Sep/24
Answered by mr W last updated on 06/Sep/24
$${say}\:{p}=\frac{{a}}{{b}},\:{q}=\frac{{b}}{{c}},\:{r}=\frac{{c}}{{a}} \\ $$$${pqr}=\mathrm{1} \\ $$$${p}+{q}+{r}=\mathrm{7} \\ $$$$\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}=\mathrm{11}\: \\ $$$$\Rightarrow\frac{{pq}+{qr}+{rp}}{{pqr}}=\mathrm{11}\:\Rightarrow{pq}+{qr}+{rp}=\mathrm{11} \\ $$$$\left({p}+{q}+{r}\right)^{\mathrm{3}} ={p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{r}^{\mathrm{3}} +\mathrm{3}\left({p}+{q}+{r}\right)\left({pq}+{qr}+{rp}\right)−\mathrm{3}{pqr} \\ $$$$\mathrm{7}^{\mathrm{3}} ={p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{r}^{\mathrm{3}} +\mathrm{3}×\mathrm{7}×\mathrm{11}−\mathrm{3}×\mathrm{1} \\ $$$$\Rightarrow{p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{r}^{\mathrm{3}} =\mathrm{7}^{\mathrm{3}} −\mathrm{3}×\mathrm{7}×\mathrm{11}+\mathrm{3}×\mathrm{1}=\mathrm{115} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} }+\frac{{b}^{\mathrm{3}} }{{c}^{\mathrm{3}} }+\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} }=\mathrm{115} \\ $$
Commented by efronzo1 last updated on 06/Sep/24
$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{pqr}\:=\:\mathrm{1}\:\mathrm{sir} \\ $$
Commented by som(math1967) last updated on 06/Sep/24
$${pqr}=\frac{{a}}{{b}}×\frac{{b}}{{c}}×\frac{{c}}{{a}}=\mathrm{1} \\ $$
Commented by Tawa11 last updated on 06/Sep/24
$$\mathrm{Sir}\:\mathrm{mrW}, \\ $$$$\mathrm{Someone}\:\mathrm{asked}\:\mathrm{question}\:\:\mathrm{211340}\:\:\mathrm{and}\:\mathrm{answered}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{reposted}\:\mathrm{because}\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{see}\:\mathrm{your}\:\mathrm{approach}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{love}\:\mathrm{to}\:\mathrm{see}\:\mathrm{your}\:\mathrm{approach}\:\mathrm{to}\:\mathrm{question}\:\mathrm{I}\:\mathrm{like}\:\mathrm{to}\:\mathrm{learn}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$