3
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What are the steps to simplify

$(1+(\frac{1}{2}(x^3-\frac{1}{x^3})^2))^ \frac{1}{2}$

to

$\frac{1}{2}(x^3+\frac{1}{x^3})$ ?

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  • $\begingroup$ Did you perhaps mean $(1+(\frac{1}{2}(x^3-\frac{1}{x^3}))^2)^ \frac{1}{2}$ $\endgroup$ – robjohn Mar 22 '14 at 10:59
1
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Hint: Try using the identities $(a-b)^2=a^2-2ab+b^2$ and $(a+b)^2=a^2+2ab+b^2$ (which are actually the same identity).

Be careful: check whether the result holds for $x\lt0$.

I am assuming you meant $$ \left(1+\left(\frac12\left(x^3-\frac1{x^3}\right)\right)^2\right)^{1/2} $$

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0
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If what you have is $$ \sqrt{ 1 + \frac{1}{2}(x^3 - \frac{1}{x^3})^2} $$

$$ \sqrt{ 1 + \frac{(x^6 - 1)^2}{2x^6}} = \sqrt{ \frac{x^{12 } + 1}{2x^6}} = \sqrt{ \frac{1}{2}(x^6 + \frac{1}{x^6})}$$

So they are not equal.

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  • 1
    $\begingroup$ Indeed. It appears that the location of the square is off. I assume this because otherwise one set of parentheses would be unnecessary. $\endgroup$ – robjohn Mar 22 '14 at 10:58
  • $\begingroup$ Ahh , then we can get the result. $\endgroup$ – neofoxmulder Mar 23 '14 at 1:24

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