Question Number 63381 by minh2001 last updated on 03/Jul/19
$${solve}\:{for}\:{both}\:{x}\:{and}\:{n} \\ $$$${in}\:{equation}:\:{x}^{{n}} =\mathrm{216}\:{in}\:{all} \\ $$$${part}\:{of}\:{integer} \\ $$$$\mathscr{A}\:\underset{{n}=\mathrm{3}} {\overset{{x}=\mathrm{6}} {\left\{}}\right. \\ $$$$\mathscr{B}\underset{{n}=\mathrm{4}} {\overset{{x}=\mathrm{5}} {\left\{}}\right. \\ $$$$\mathscr{C}\underset{{n}=\mathrm{5}} {\overset{{x}=\mathrm{4}} {\left\{}}\right. \\ $$$$\mathscr{D}\:\underset{{n}=\mathrm{6}} {\overset{{x}=\mathrm{3}} {\left\{}}\right. \\ $$
Commented by mr W last updated on 03/Jul/19
$$\mathrm{216}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{3}} =\left(\mathrm{2}×\mathrm{3}\right)^{\mathrm{3}} =\mathrm{6}^{\mathrm{3}} \\ $$$$\mathrm{216}=\mathrm{216}^{\mathrm{1}} \\ $$$${solutions}\:{are}\:{x}=\mathrm{6},\:{n}=\mathrm{3}\:{or}\:{x}=\mathrm{216},\:{n}=\mathrm{1} \\ $$
Commented by minh2001 last updated on 04/Jul/19
$${You}'{re}\:{right}\:{but}\:{i}\:{just}\:{need} \\ $$$${only}\:{one}\:{answer}\:.{You}'{ve} \\ $$$${choosen}\:{an}\:{answer}\:\mathscr{A},{haven}'{t}\:{you}? \\ $$