Question

answer fast. plz answer fast

Answer 1

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$\red{\underline{SOLUTION}}$ ➡

$Given \: ,\\ \\ z {}^{2} + \frac{1}{z {}^{2} } = 14 \\ \\ To \: find : z {}^{3} + \frac{1}{z {}^{3} } \\ \\ Now, \\ \\ \\ z {}^{2} + \frac{1}{ z{}^{2} } = 14 \\ \\ = > z {}^{2} + (\frac{1}{z} ) {}^{2} + 2 \times z \times \frac{1}{z} - 2 \times z \times \frac{1}{z} = 14 \\ \\ = > (z + \frac{1}{z} ) {}^{2} - 2 = 14 \\ \\ = > (z + \frac{1}{z} ) {}^{2} = 14 + 2 \\ \\ = > (z + \frac{1}{z} ) {}^{2} = 16 \\ \\ = > (z + \frac{1}{z} ) {}^{2} = (4) {}^{2} \\ \\ = > z + \frac{1}{z} = 4 \\ \\ \\ \\ So \: ,\\ \\ \\ (z + \frac{1}{z} ) {}^{3} = (4) {}^{3} \\ \\ = > (z + \frac{1}{z} ) {}^{3} = 64 \\ \\ = > z {}^{3} + (\frac{1}{z} ) {}^{3} + 3(z + \frac{1}{z} ) = 64 \\ \\ = > z {}^{3} + \frac{1}{z {}^{3} } + 3 \times 4 = 64 \\ \\ = > z {}^{3} + \frac{1}{z {}^{3} } + 12 = 64 \\ \\ = > z {}^{3} + \frac{1}{z {}^{3} } = 64 - 12 \\ \\ = > z {}^{3} + \frac{1}{z {}^{3} } = 52$

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