x $ ^7+x^4+i(x^3+1)=0 \\ \; \\ \; \\ \therefore x^4 (x^3+1)+i(x^3+1)=0,i.e.(x^4+i)(x^3+1)=0 \\ \; \\ \; \\ \therefore (x^4+i)=0 \; and \; (x^3+1)=0 \\ \; \\ $

Consider $x^4+i=0, \;i.e. \; x^4=-i, \; i.e. \; x^4=cos\dfrac{3\pi}{2}+i sin\dfrac{3\pi}{2}$

In General, $ x^4=cos\bigg( 2n\pi + \dfrac{3\pi}{2} \bigg)+i sin \bigg( 2n\pi + \dfrac{3\pi}{2} \bigg) $

$ \therefore x \;=\; \Bigg[ cos\bigg( 2n\pi + \dfrac{3\pi}{2} \bigg)+i sin \bigg( 2n\pi + \dfrac{3\pi}{2} \bigg) \Bigg]^{\frac{1}{4}} \\ \; \\ \; \\ \; \\ \therefore x \;=\; \Bigg[ cos\bigg( \dfrac{2n\pi}{4} + \dfrac{3\pi}{8} \bigg)+i sin \bigg( \dfrac{2n\pi}{4} + \dfrac{3\pi}{8} \bigg) \Bigg] $ where n = 0,1,2,3....(By DeMoivre's Theorem)

$ \therefore x_1 \;=\; \bigg[ cos\Big( \dfrac{3\pi}{8} \Big) + sin\Big( \dfrac{3\pi}{8} \Big)\bigg] \\ \; \\ \; \\ \; \\ x_2 \;=\; \bigg[ cos\Big( \dfrac{7\pi}{8} \Big) + sin\Big( \dfrac{7\pi}{8} \Big)\bigg] \\ \; \\ \; \\ \; \\ x_3 \;=\; \bigg[ cos\Big( \dfrac{11\pi}{8} \Big) + sin\Big( \dfrac{11\pi}{8} \Big)\bigg] \\ \; \\ \; \\ \; \\ x_4 \;=\; \bigg[ cos\Big( \dfrac{15\pi}{8} \Big) + sin\Big( \dfrac{15\pi}{8} \Big)\bigg] \\ \; \\ \; \\ \; \\ $

Now,consider $x^3+1=0, \; i.e. \; x^3=-1, \; i.e. \; x^3=cos\pi +i sin\pi⁡$

In General,$x^3=cos⁡(2n\pi+\pi)+i sin⁡(2n\pi+\pi)$

$ x \;=\; \Bigg[ cos\bigg( 2n\pi + \pi \bigg)+i sin \bigg( 2n\pi + \pi \bigg) \Bigg]^{\frac{1}{3}} $ where n=0,1,2

$ x \;=\; \Bigg[ cos\bigg( \dfrac{2n\pi}{3} + \dfrac{\pi}{3} \bigg)+i sin \bigg( \dfrac{2n\pi}{3} + \dfrac{\pi}{3} \bigg) \Bigg] $ ......(By DeMoivre's Theorem)

$ \therefore x_5 \;=\; \bigg[ cos\Big( \dfrac{\pi}{3} \Big) + sin\Big( \dfrac{\pi}{3} \Big)\bigg] \;=\; \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i \\ \; \\ \; \\ \; \\ x_6 \;=\; \bigg[ cos\Big(\pi \Big) + sin\Big( \pi \Big)\bigg] \;=\; -1 \\ \; \\ \; \\ \; \\ x_7 \;=\; \bigg[ cos\Big( \dfrac{5\pi}{3} \Big) + sin\Big( \dfrac{5\pi}{3} \Big)\bigg] \;=\; \dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i \\ \; \\ \; \\ \; \\ $