Question Number 197708 by karowan last updated on 26/Sep/23
$$ \\ $$$${x}^{{a}} ={x}^{{a}+\mathrm{4}} \:\:\mathrm{where}\:{a}\in\mathbb{Z} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x}\:\mathrm{by}\:\mathrm{showing}\:\mathrm{steps} \\ $$
Answered by Frix last updated on 26/Sep/23
$${x}^{{a}+\mathrm{4}} −{x}^{{a}} =\mathrm{0} \\ $$$${x}^{{a}} \left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${a}>\mathrm{0}\wedge{x}=\mathrm{0} \\ $$$${x}=\pm\mathrm{1} \\ $$$${x}=\pm\mathrm{i} \\ $$
Commented by karowan last updated on 26/Sep/23
correct