Question Number 211636 by BaliramKumar last updated on 15/Sep/24
Answered by Rasheed.Sindhi last updated on 15/Sep/24
$$\blacktriangleright{A}={x}^{\mathrm{8}{k}+\mathrm{3}} +{x}^{\mathrm{8}{k}+\mathrm{6}} +{x}^{\mathrm{8}{k}+\mathrm{9}} +{x}^{\mathrm{8}{k}+\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:={x}^{\mathrm{8}{k}+\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +{x}^{\mathrm{9}} \right) \\ $$$$\blacktriangleright{B}=\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\left(\mathrm{1}+{x}^{\mathrm{6}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +{x}^{\mathrm{9}} \\ $$$$\frac{{A}}{{B}}={x}^{\mathrm{8}{k}+\mathrm{3}} \\ $$
Commented by BaliramKumar last updated on 15/Sep/24
$${thanks} \\ $$