Question Number 80161 by peter frank last updated on 01/Feb/20
$${Find}\:{the}\:{relation}\:{between} \\ $$$${q}\:{and}\:{r}\:\:{so}\:\:{that} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}\:{is}\:{a}\:{perfect} \\ $$$${cube}\:{for}\:{all}\:\:{value}\:{of}\:{x} \\ $$
Commented by peter frank last updated on 31/Jan/20
$${yes}\:{sir} \\ $$
Commented by mr W last updated on 31/Jan/20
$${you}\:{mean}\:{x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}\:? \\ $$
Commented by mr W last updated on 31/Jan/20
$$\Rightarrow{q}=\mathrm{3}{p}^{\mathrm{2}} \\ $$$$\Rightarrow{r}={p}^{\mathrm{3}} \\ $$$${then}\:\:{x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}=\left({x}+{p}\right)^{\mathrm{3}} \\ $$
Commented by mr W last updated on 31/Jan/20
$${is}\:{this}\:{what}\:{you}\:{meant}? \\ $$
Commented by peter frank last updated on 31/Jan/20
$${i}\:{dont}\:{understand}\:{the}\: \\ $$$${question}\:{sir}.{what}\:{if} \\ $$$${the}\:{equation}\:{remain} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}+{qx}+{r}\:{is}\:{it}\: \\ $$$${possible}\:{find}\:{relation} \\ $$$${r}\:{and}\:{q}? \\ $$
Commented by mr W last updated on 31/Jan/20
$${since}\:{q}=\mathrm{3}{p}^{\mathrm{2}} \:{and}\:{r}={p}^{\mathrm{3}} \:{you}\:{can}\:{say} \\ $$$$\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} ={r}^{\mathrm{2}} \:{or}\:{q}^{\mathrm{3}} =\mathrm{27}{r}^{\mathrm{2}} . \\ $$
Commented by MJS last updated on 31/Jan/20
$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}=\left({x}−{z}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${z}+{p}=\mathrm{0}\wedge\mathrm{3}{z}^{\mathrm{2}} −{q}=\mathrm{0}\wedge{z}^{\mathrm{3}} +{r}=\mathrm{0} \\ $$$${z}=−{p} \\ $$$${q}=\mathrm{3}{p}^{\mathrm{2}} \\ $$$${r}={p}^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\frac{{q}}{{r}}=\frac{\mathrm{3}}{{p}} \\ $$
Commented by peter frank last updated on 01/Feb/20
$${thank}\:{you}\:@\:{mr}\:{w}\:@{MJS} \\ $$
Commented by mr W last updated on 01/Feb/20
$${MJS}\:{sir}: \\ $$$${i}\:{think}\:\frac{{q}}{{r}}=\frac{\mathrm{3}}{{p}}\:{doesn}'{t}\:{ensure}\:{that} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}=\left({x}−{z}\right)^{\mathrm{3}} .\:{e}.{g}.\:{with} \\ $$$${p}=\mathrm{2},\:{q}=\pm\mathrm{3},\:{r}=\pm\mathrm{2}\:{we}\:{habe} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}={x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} \pm\mathrm{3}{x}\pm\mathrm{2}.\:{but} \\ $$$${both}\:\neq\left({x}−{z}\right)^{\mathrm{3}} . \\ $$
Commented by MJS last updated on 01/Feb/20
$$\mathrm{we}\:\mathrm{need}\:\mathrm{also}\:{q}=\mathrm{3}{p}^{\mathrm{2}} \wedge{r}={p}^{\mathrm{3}} \\ $$$${p}=\mathrm{2}\:\Rightarrow\:{q}=\mathrm{12}\wedge{r}=\mathrm{8};\:{z}=−\mathrm{2} \\ $$$${x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{8}=\left({x}+\mathrm{2}\right)^{\mathrm{3}} \\ $$