If-x-3-1-y-3-y-3-1-z-3-1-Find-xyz-2025-1-

Question Number 154316 by mathdanisur last updated on 17/Sep/21

If x^{3} + \frac{1}{y^{3}} = y^{3} + \frac{1}{z^{3}} = 1

Find {(xyz)}^{2025} - 1 = ?

Answered by Rasheed.Sindhi last updated on 17/Sep/21

x^{3} + \frac{1}{y^{3}} = y^{3} + \frac{1}{z^{3}} = 1; {(xyz)}^{2025} - 1 = ?

{\begin{cases} x^{3} = 1 - \frac{1}{y^{3}} = \frac{y^{3} - 1}{y^{3}} \\ z^{3} = \frac{1}{1 - y^{3}} = - \frac{1}{y^{3} - 1} \end{cases}

▸ {(xyz)}^{2025} - 1 = {(x^{3} y^{3} z^{3})}^{675} - 1

= {((\frac{y^{3} - 1}{y^{3}}) (y^{3}) (- \frac{1}{y^{3} - 1}))}^{675} - 1

= {(- 1)}^{675} - 1 = - 1 - 1 = - 2

Commented by mathdanisur last updated on 17/Sep/21

Nice Ser, thank you