If #x+1/x=2# then what is #(x+1/x)^3#?
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To find ( (x + \frac{1}{x})^3 ), first expand ( (x + \frac{1}{x})^3 ) using the binomial theorem or by direct cubing:
[ (x + \frac{1}{x})^3 = x^3 + 3x^2(\frac{1}{x}) + 3x(\frac{1}{x})^2 + (\frac{1}{x})^3 ]
Given that ( x + \frac{1}{x} = 2 ), we can substitute this value into the expression:
[ (2)^3 = x^3 + 3x^2(\frac{1}{x}) + 3x(\frac{1}{x})^2 + (\frac{1}{x})^3 ]
Since ( x + \frac{1}{x} = 2 ), ( x^2 + 2x(\frac{1}{x}) + (\frac{1}{x})^2 = 4 ):
[ (2)^3 = x^3 + 3(2) + 3(2) + (\frac{1}{x})^3 ]
[ 8 = x^3 + 6 + \frac{3}{x} + \frac{1}{x^3} ]
Rearrange the terms:
[ x^3 + \frac{1}{x^3} = 8 - 6 = 2 ]
Therefore, ( (x + \frac{1}{x})^3 = 2 ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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