8(3^2+1)(3^4+1)(3^8+1)...(3^32+1)+1已知:a^2-3a-1=0,a^2+1/a^2=?

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8(3^2+1)(3^4+1)(3^8+1)...(3^32+1)+1已知:a^2-3a-1=0,a^2+1/a^2=?
8(3^2+1)(3^4+1)(3^8+1)...(3^32+1)+1
已知:a^2-3a-1=0,a^2+1/a^2=?

8(3^2+1)(3^4+1)(3^8+1)...(3^32+1)+1已知:a^2-3a-1=0,a^2+1/a^2=?
8(3^2+1)(3^4+1)(3^8+1)...(3^32+1)+1
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...(3^32+1)+1
=(3^4-1)(3^4+1)(3^8+1)...(3^32+1)+1
=(3^8-1)(3^8+1)...(3^32+1)+1
=.
=(3^32-1)(3^32+1)+1
=(3^64-1)+1
=3^64
已知:a^2-3a-1=0,a^2+1/a^2=?
由a^2-3a-1=0知a≠0
所以a-1/a=3
两边平方得:
a^2-2+1/a^2=9
所以
a^2+1/a^2=9+2=11

第一题:乘以(3^2-1),连续用平方差公式,再除回来
原式=8(3^64-1)/(3^2-1) +1 =3^64
第二题:a-3-1/a=0
a-1/a=3
两边平方
a^2 - 2 + 1/a^2 =9
故答案为11