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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  + (1/y^3 ) = y^3  + (1/z^3 ) = 1  Find  (xyz)^(2025)  - 1 = ?
$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }\:=\:\mathrm{y}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }\:=\:\mathrm{1} \\ $$$$\mathrm{Find}\:\:\left(\mathrm{xyz}\right)^{\mathrm{2025}} \:-\:\mathrm{1}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 17/Sep/21
  x^3  + (1/y^3 ) = y^3  + (1/z^3 ) = 1; (xyz)^(2025)  - 1 = ?   { ((x^3 =1−(1/y^3 )=((y^3 −1)/y^3 ))),((z^3 =(1/(1−y^3 ))=−(1/(y^3 −1)))) :}   ▶(xyz)^(2025)  - 1=(x^3 y^3 z^3 )^(675) −1  =((((y^3 −1)/y^3 ))(y^3 )(−(1/(y^3 −1))))^(675) −1  =(−1)^(675) −1=−1−1=−2
$$ \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }\:=\:\mathrm{y}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }\:=\:\mathrm{1};\:\left(\mathrm{xyz}\right)^{\mathrm{2025}} \:-\:\mathrm{1}\:=\:? \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }=\frac{\mathrm{y}^{\mathrm{3}} −\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }}\\{\mathrm{z}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} −\mathrm{1}}}\end{cases} \\ $$$$\:\blacktriangleright\left(\mathrm{xyz}\right)^{\mathrm{2025}} \:-\:\mathrm{1}=\left(\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} \mathrm{z}^{\mathrm{3}} \right)^{\mathrm{675}} −\mathrm{1} \\ $$$$=\left(\left(\frac{\cancel{\mathrm{y}^{\mathrm{3}} −\mathrm{1}}}{\cancel{\mathrm{y}^{\mathrm{3}} }}\right)\left(\cancel{\mathrm{y}^{\mathrm{3}} }\right)\left(−\frac{\mathrm{1}}{\cancel{\mathrm{y}^{\mathrm{3}} −\mathrm{1}}}\right)\right)^{\mathrm{675}} −\mathrm{1} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{675}} −\mathrm{1}=−\mathrm{1}−\mathrm{1}=−\mathrm{2} \\ $$
Commented by mathdanisur last updated on 17/Sep/21
Nice Ser, thank you
$$\mathrm{Nice}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$

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