Question Number 29989 by naka3546 last updated on 14/Feb/18
Answered by MJS last updated on 14/Feb/18
$${a}=\mathrm{2}\:{b}={c}=\mathrm{0} \\ $$$${a}^{\mathrm{2018}} +{b}^{\mathrm{2018}} +{c}^{\mathrm{2018}} =\mathrm{2}^{\mathrm{2018}} \approx\mathrm{3}×\mathrm{10}^{\mathrm{607}} \\ $$