Question Number 100767 by Rio Michael last updated on 28/Jun/20
$$\:\mathrm{sketch}\:\:{x}^{\mathrm{2}} \:=\:{y}^{\mathrm{3}} \:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\mathrm{16} \\ $$$$\mathrm{and}\:\mathrm{hence}\:\mathrm{solve}\:\left({x}^{\mathrm{2}} −{y}^{\mathrm{3}} \right)\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} −\mathrm{16}\right)\:\geqslant\mathrm{0} \\ $$
Answered by john santu last updated on 28/Jun/20
Answered by john santu last updated on 28/Jun/20
