Solution:
_______________________
We know the algebraic identity:
(x+y)(x-y) = x² - y²
________________________
Here ,
(a+1/a)(a-1/a)(a²+1/a²)(a⁴+1/a⁴)
=[(a+1/a)(a-1/a)](a²+1/a²)(a⁴+1/a⁴)
=[a²-(1/a)²](a²+1/a²)(a⁴+1/a⁴)
=[(a²-1/a²)(a²+1/a²)](a⁴+1/a⁴)
= [(a²)²-(1/a²)²](a⁴+1/a⁴)
= (a⁴-1/a⁴)(a⁴+1/a⁴)
= (a⁴)²-(1/a⁴)²
=
Therefore,
(a+1/a)(a-1/a)(a²+1/a²)(a⁴+1/a⁴) =
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Answers & Comments
Solution:
_______________________
We know the algebraic identity:
(x+y)(x-y) = x² - y²
________________________
Here ,
(a+1/a)(a-1/a)(a²+1/a²)(a⁴+1/a⁴)
=[(a+1/a)(a-1/a)](a²+1/a²)(a⁴+1/a⁴)
=[a²-(1/a)²](a²+1/a²)(a⁴+1/a⁴)
=[(a²-1/a²)(a²+1/a²)](a⁴+1/a⁴)
= [(a²)²-(1/a²)²](a⁴+1/a⁴)
= (a⁴-1/a⁴)(a⁴+1/a⁴)
= (a⁴)²-(1/a⁴)²
=![a^{8}-\frac{1}{a^{8}} a^{8}-\frac{1}{a^{8}}](https://tex.z-dn.net/?f=a%5E%7B8%7D-%5Cfrac%7B1%7D%7Ba%5E%7B8%7D%7D)
Therefore,
(a+1/a)(a-1/a)(a²+1/a²)(a⁴+1/a⁴) =![a^{8}-\frac{1}{a^{8}} a^{8}-\frac{1}{a^{8}}](https://tex.z-dn.net/?f=a%5E%7B8%7D-%5Cfrac%7B1%7D%7Ba%5E%7B8%7D%7D)
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