Question

Find the cube root of -27 and show that the sum of cube toots is equal to zero

Answer 1

$$\begin{gathered} x^3=-27 \\ {x}^3=-3^3 \\ {x}^3=3^3\left(-1\right) \\ weknow{e}^{i\left(2k+1\right)\mathrm{\pi }}=-1 \\ {x}^3=3^3{e}^{i\left(2k+1\right)\mathrm{\pi }} \\ x={\left\{3^3{e}^{i\left(2k+1\right)\mathrm{\pi }}\right\}}^{\frac{1}{3}} \\ x=3e^{i\left(\frac{\left(2\mathrm{k}+1\right)\mathrm{\pi }}{3}\right)} \\ Putk=0,x=3e^{i\left(\frac{\mathrm{\pi }}{3}\right)}=3\left(\cos \frac{\mathrm{\pi }}{3}+i\sin \frac{\mathrm{\pi }}{3}\right)=3\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) \\ Putk=1,x=3e^{i\left(\frac{3\mathrm{\pi }}{3}\right)}=3e^{i\mathrm{\pi }}=3\left(\cos \mathrm{\pi }+i\sin \mathrm{\pi }\right)=3\left(-1+0\right)=-3 \\ Putk=2,x=3e^{i\left(\frac{5\mathrm{\pi }}{3}\right)}=3\left(\cos \frac{5\mathrm{\pi }}{3}+i\sin \frac{5\mathrm{\pi }}{3}\right)=3\left(\cos \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)+i\sin \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right)=3\left(\cos \frac{\mathrm{\pi }}{3}-i\sin \frac{\mathrm{\pi }}{3}\right)=3\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right) \\ Thesearethreecuberootsof-27 \\ 3\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+\left(-3\right)+3\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right) \\ =3\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}+\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)+\left(-3\right) \\ =3\left(\frac{1}{2}+\frac{1}{2}\right)+\left(-3\right) \\ =3-3 \\ =0 \end{gathered}$$