Question Number 157694 by ajfour last updated on 26/Oct/21
$$\:\:\:{x}^{\mathrm{3}} ={x}+{c}\:\:\:\:\:;\:\:\:\:\mathrm{0}<{c}\leqslant\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${find}\:{x},\:{without}\:{trigonometric} \\ $$$${cubic}\:{formula}. \\ $$
Answered by ajfour last updated on 26/Oct/21
