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Factor completely…..c^92 - t^92

User A Salim
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1 Answer

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Okay, here we have this, considering the provided information, we are going to factor it completely, so we obtain:


\begin{gathered} c^{\mleft\{92\mright\}}-t^{\mleft\{92\mright\}} \\ =\mleft(c^(46)\mright)^2-\mleft(t^(46)\mright)^2 \\ =\mleft(c^(46)+t^(46)\mright)\mleft(c^(46)-t^(46)\mright) \\ =(\mleft(c^2\mright)^(23)+\mleft(t^2\mright)^(23))(\mleft(c^(23)\mright)^2-\mleft(t^(23)\mright)^2) \\ =\mleft(c^2+t^2\mright)\mleft(c^(44)-t^2c^(42)+t^4c^(40)-t^6c^(38)+t^8c^(36)-t^(10)c^(34)+t^(12)c^(32)-t^(14)c^(30)+t^(16)c^(28)-t^(18)c^(26)+t^(20)c^(24)-t^(22)c^(22)+t^(24)c^(20)-t^(26)c^(18)+t^(28)c^(16)-t^(30)c^(14)+t^(32)c^(12)-t^(34)c^(10)+t^(36)c^8-t^(38)c^6+t^(40)c^4-t^(42)c^2+t^(44)\mright)\mleft(c^(46)-t^(46)\mright) \\ =\mleft(c^2+t^2\mright)\mleft(c^(44)-t^2c^(42)+t^4c^(40)-t^6c^(38)+t^8c^(36)-t^(10)c^(34)+t^(12)c^(32)-t^(14)c^(30)+t^(16)c^(28)-t^(18)c^(26)+t^(20)c^(24)-t^(22)c^(22)+t^(24)c^(20)-t^(26)c^(18)+t^(28)c^(16)-t^(30)c^(14)+t^(32)c^(12)-t^(34)c^(10)+t^(36)c^8-t^(38)c^6+t^(40)c^4-t^(42)c^2+t^(44)\mright)\mleft(c+t\mright)\mleft(c^(22)-tc^(21)+t^2c^(20)-t^3c^(19)+t^4c^(18)-t^5c^(17)+t^6c^(16)-t^7c^(15)+t^8c^(14)-t^9c^(13)+t^(10)c^(12)-t^(11)c^(11)+t^(12)c^(10)-t^(13)c^9+t^(14)c^8-t^(15)c^7+t^(16)c^6-t^(17)c^5+t^(18)c^4-t^(19)c^3+t^(20)c^2-t^(21)c+t^(22)\mright)\mleft(c-t\mright)\mleft(c^(22)+tc^(21)+t^2c^(20)+t^3c^(19)+t^4c^(18)+t^5c^(17)+t^6c^(16)+t^7c^(15)+t^8c^(14)+t^9c^(13)+t^(10)c^(12)+t^(11)c^(11)+t^(12)c^(10)+t^(13)c^9+t^(14)c^8+t^(15)c^7+t^(16)c^6+t^(17)c^5+t^(18)c^4+t^(19)c^3+t^(20)c^2+t^(21)c+t^(22)\mright) \end{gathered}

User Keelerm
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