Question Number 192173 by universe last updated on 10/May/23
$${prove}\:{that} \\ $$$$\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} \:=\:\left({x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{2}} {a}^{\mathrm{2}} +{a}^{\mathrm{4}} \right)^{\mathrm{2}} +\left(\mathrm{4}{x}^{\mathrm{3}} {a}−\mathrm{4}{xa}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$
Answered by mehdee42 last updated on 10/May/23
$${right}\:{side}=\left(\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{4}} −\mathrm{8}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{4}} +\mathrm{8}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left(\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} \:\checkmark \\ $$$$ \\ $$