Question

(ii) x^7+ 1 = 0 solve by de moivre theorem​

Answer 1

Step-by-step explanation:

For an odd-order polynomial, there must be at least one real root. For (1), a real root is x=−1, and for (2) a real root is x=+1. You may divide out those roots if you wish.

For complex roots, x=|x| (cosθ±i sinθ).

EDIT: I highly recommend dividing out the real root of each equation, and then factoring the resulting 6th-order polynomial into irreducible quartic and quadratic factors.

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