Question #158841

x+1/x=5, then find the value of x^16+1/x^16 ?


1
Expert's answer
2021-01-29T04:44:58-0500

Given

x+1/x=5x+1/x=5

Tahing square on both sides we have

(x+1/x)2=52(x+1/x)^2=5^2

solving the bracket by using formula (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2

we get

x2+2(x)(1/x)+1/x2=25,x^2+2(x)(1/x)+1/x^2=25,

x2+1/x2=252x^2+1/x^2=25-2

=x2+1/x2=23=x^2+1/x^2=23

again taking aquare on both sides we have ,

(x2+1/x2)2=232(x^2+1/x^2)^2=23^2

=x4+(2)(x2)(1/x2)+1/x4=529,=x^4+(2)(x^2)(1/x^2)+1/x^4=529,

x4+1/x4=5292x^4+1/x^4=529-2


=x4+1/x4=527=x^4+1/x^4=527


taking aquare on both sides we have

=(x4+1/x4)2=(527)2=(x^4+1/x^4)^2=(527)^2

=x8+(2)(x4)(1/x4)+1/x8=277729,=x^8+(2)(x^4)(1/x^4)+1/x^8=277729,

x8+(2)+1/x8=2777292x^8+(2)+1/x^8=277729-2

=x8+1/x8=277727=x^8+1/x^8=277727

taking aquare on both sides we have

=(x8+1/x8)2=(277727)2=(x^8+1/x^8)^2=(277727)^2

=x16+(2)(x8)(1/x8)+1/x16=(277727)2,=x^{16}+(2)(x^8)(1/x^8)+1/x^{16}=(277727)^2,

x16+1/x16=(277727)22x^{16}+1/x^{16}=(277727)^2-2





Answer:

x16+1/x16=(277727)22x^{16}+1/x^{16}=(277727)^2-2


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