different one

Algebra Level 4

x 4 x 3 + x 2 x + 1 = 0 x^{4}-x^{3}+x^{2}-x+1=0 has "a" as one root.then value of a 20 a 15 + a 10 a 5 + 1 a^{20}-a^{15}+a^{10}-a^{5}+1 is ??


The answer is 5.

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1 solution

Sai Aryanreddy
Nov 4, 2015

AS they gave "a" as one of the root then, a 4 a 3 + a 2 a + 1 = 0 a^{4}-a^{3}+a^{2}-a+1=0 CONSIDER, a 5 + 1 a^{5}+1 a 5 + 1 = ( a + 1 ) ( a 4 a 3 + a 2 a + 1 = 0 ) a^{5}+1=(a+1)(a^{4}-a^{3}+a^{2}-a+1=0) a 5 + 1 = ( a + 1 ) ( 0 ) a^{5}+1=(a+1)(0) a 5 = 1 a^{5}=-1 substitute this value in a 20 a 15 + a 10 a 5 + 1 a^{20}-a^{15}+a^{10}-a^{5}+1 we get it as ( 1 ) 4 ( 1 ) 3 + ( 1 ) 2 ( 1 ) + 1 = 1 + 1 + 1 + 1 + 1 = 5 (-1)^{4}-(-1)^{3}+(-1)^{2}-(-1)+1=1+1+1+1+1=\boxed{5}

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