Question

If z Square +1 upon z square is equal to 14 find the value of z cube plus one upon z cube

Answer 1

$given \: \: {z}^{2} + \frac{1}{ {z}^{2} } = 14 \\ {(z + \frac{1}{z} )}^{2} = {z}^{2} + \frac{1}{ {z}^{2} } + 2 \\ = 14 + 2 = 16 \\ therefore \: \\ z + \frac{1}{z} = \sqrt{16} = 4 \: \: or \: \: - 4$
${z}^{3} + \frac{1}{ {z}^{3} } = {(z + \frac{1}{z} )}^{3} - 3 (z + \frac{1}{z}) \\ = {4}^{3} - 3 \times 4 \: \: or \: \: ( { - 4)}^{3} - 3 \times - 4 \\ = 52 \: \: or \: \: - 52$

Answer 2

ANSWER:-
The answer is 52/-52 and the solution is provided in the attachment.
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