Question Number 201814 by Euclid last updated on 13/Dec/23
Answered by sajitha last updated on 13/Dec/23
$$\mathrm{4}^{{x}^{\mathrm{3}} } =\mathrm{4} \\ $$$${x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left\{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right\}=\mathrm{0} \\ $$$${x}=\mathrm{1},{x}=\frac{−\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}\:}{i}}{\mathrm{2}} \\ $$