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Bunuel
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Bunuel
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A straightforward way to solve this is to equate \(n^6\) to the answer choices and see if they equal \(n^4 + n^2 = 1\)

Intuitively, the lower options looked more suitable than the top ones so I started from the bottom and worked my way up.

\(2n^2 + 1=n^6\) ­

\(n^6 - 2n^2 = 1\) ­

\(n^2(n^4 - 2) = 1\) ­

Here I rearranged \(n^4 + n^2 = 1\) to \(n^4 = 1 - n^2\)​​​​​​​ and plugged in \(n^4\)

\(n^2(1 - n^2 - 2) = 1\) ­
​​​​​​​
\(- n^2 - n^4 = 1\) ­or \(n^2 + n^4 = - 1\) Doesn't work

We try again with the next value

\(2n^2 - 1=n^6\) ­

\(n^6 - 2n^2 = - 1\) ­

\(n^2(n^4 - 2) = - 1\) ­

\(n^2(1 - n^2 - 2) = - 1\) ­

\(- n^2 - n^4 = - 1\) or \(n^2 + n^4 = 1\) which works

The answer is D
Bunuel
­If \(n^4 + n^2 = 1\), then which of the following must equal to \(n^6\)?

A. \(n^2 - 1\)

B. \(n^2 + 1\)

C. \(2n^2\)

D. \(2n^2 - 1\)

E. \(2n^2 + 1\) ­

 


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\(n^4+n^2=1 => n^6 + n^4 = n^2 => n^6 = n^2 +n^2 - n^2 - n^4 = 2*n^2-1 \)­

The right answer is D­
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