Question

${z}^{2} + \frac{1}{ {z}^{2} } = 14 \: find \: {z}^{3} + \frac{1}{ {z}^{3} }$
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

${z}^{2} + \frac{1}{ {z}^{2} } = 14 \\ = {z}^{2} + \frac{1}{ {z}^{2} } + 2= 16 \\ (z + \frac{1}{z} )^{2} = {4}^{2} \\ = z + \frac{1}{z} = + - 4 \\ {z}^{3} + \frac{1}{ {z}^{3} } = (z + \frac{1}{z} )( {z}^{2} + \frac{1}{ {z}^{2} } - 1) \\ = + - 4(14 - 1) \\ = + - 4(13) \\ = + - 52$