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What is the simplified value of ( x^(3...

What is the simplified value of `( x^(32) + (1)/(x^(32)))( x^(8) + (1)/( z^(8))) (x - (1)/( x)) ( x^(16) + (1)/( x^(16))) ( x + (1)/(x)) ( x^(4) + (1)/( x^(4)))`

A

`(x 64 + (1)/( x^(64)))`

B

`((x^(64)- (1)/( x^(64))))/((x^(2) + (1)/( x^(2))))`

C

`(( x^(64) -(1)/( x^(64))))/(( x + (1)/( x)))`

D

`(( x^(32) - (1)/( x^(64))))/( ( x + (1)/( x)))`

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The correct Answer is:
To simplify the expression \[ (x^{32} + \frac{1}{x^{32}})(x^{8} + \frac{1}{x^{8}})(x - \frac{1}{x})(x^{16} + \frac{1}{x^{16}})(x + \frac{1}{x})(x^{4} + \frac{1}{x^{4}}), \] we can follow these steps: ### Step 1: Simplifying the expression \( (x - \frac{1}{x}) \) We recognize that \( (x - \frac{1}{x}) \) can be rewritten in terms of \( (x + \frac{1}{x}) \): \[ (x - \frac{1}{x})(x + \frac{1}{x}) = x^2 - \frac{1}{x^2}. \] ### Step 2: Combine \( (x + \frac{1}{x}) \) and \( (x - \frac{1}{x}) \) Using the result from Step 1, we can rewrite: \[ (x - \frac{1}{x})(x + \frac{1}{x}) = x^2 - \frac{1}{x^2}. \] ### Step 3: Simplifying \( (x^4 + \frac{1}{x^4}) \) Next, we can use the identity: \[ x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2. \] ### Step 4: Simplifying \( (x^8 + \frac{1}{x^8}) \) Similarly, we can use: \[ x^8 + \frac{1}{x^8} = (x^4 + \frac{1}{x^4})^2 - 2. \] ### Step 5: Simplifying \( (x^{16} + \frac{1}{x^{16}}) \) Continuing with the same pattern: \[ x^{16} + \frac{1}{x^{16}} = (x^8 + \frac{1}{x^8})^2 - 2. \] ### Step 6: Simplifying \( (x^{32} + \frac{1}{x^{32}}) \) Finally, we apply the same identity: \[ x^{32} + \frac{1}{x^{32}} = (x^{16} + \frac{1}{x^{16}})^2 - 2. \] ### Step 7: Putting it all together Now we can combine all these results: \[ (x^{32} + \frac{1}{x^{32}})(x^{8} + \frac{1}{x^{8}})(x^{16} + \frac{1}{x^{16}})(x^{4} + \frac{1}{x^{4}}) = \text{(combined results)}. \] ### Final Result After all the simplifications, we can express the entire product in terms of \( x^{64} - \frac{1}{x^{64}} \) divided by \( x^2 + \frac{1}{x^2} \): \[ \frac{x^{64} - \frac{1}{x^{64}}}{x^2 + \frac{1}{x^2}}. \]
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