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**cmf0106** In the difference of squares it can come in the form of $\displaystyle x^n-c^n$ where n is an even number. The factorization is $\displaystyle x^n-c^n=(x^{\frac{n}{2}}-c^{\frac{n}{2}})(x^{\frac{n}{2}}+c^{\frac{n}{2}})$

However in the following problem I am lost at a particular step. Namely, how $\displaystyle \frac{1}{64}$ turns into $\displaystyle (\frac{1}{2})^6$

$\displaystyle x^6-\frac{1}{64}=x^6-(\frac{1}{2})^6=(x^3-\frac{1}{2})(x^3+\frac{1}{2})$

If someone could walk me through this it would be greatly appreciated,

Many Thanks