Ti-83 difference of squares

**allyourbass2212** · August 4th 2009, 08:20 AM

$x^n-c^n=(x^\frac{n}{2}-c\frac{n}{2})$

I like to use the calculator to double check myself, but for some reason I cannot produce the correct answer for the problem below.

$x^8-\frac{1}{16}=(x^4+\frac{1}{4})(x^4+\frac{1}{4})$
$=(x^2-\frac{1}{2})(x^2+\frac{1}{2})(x^4+\frac{1}{4})$

The first step I attempt to get $(x^4+\frac{1}{4})(x^4+\frac{1}{4})$ by putting this into the calculator $(\frac{1}{16})^{(\frac{1}{4})}$ which results in .5 or 1/2. How do I keep on missing the first step to the solution which is the 1/4?

**pickslides** · August 4th 2009, 02:16 PM

Originally Posted by allyourbass2212

$x^8-\frac{1}{16}=(x^4+\frac{1}{4})(x^4+\frac{1}{4})$

I don't think this is correct. I believe it is $x^8-\frac{1}{16}=(x^4-\frac{1}{4})(x^4+\frac{1}{4})$

**allyourbass2212** · August 4th 2009, 08:22 PM

Sorry that was a typo, and my question is where does the

come from, and from what operation exactly in the calculator?

**cmf0106** · August 4th 2009, 08:40 PM

Hello Allyourbass,

What squared is 16? Simply input $\frac{1}{16}^{\frac{1}{2}}$ into your calculator and it will yield $\frac{1}{4}$

**pickslides** · August 4th 2009, 09:52 PM

Originally Posted by allyourbass2212

Sorry that was a typo, and my question is where does the

come from, and from what operation exactly in the calculator?

this comes from the difference of 2 squares rule.

$a^2-b^2 = (a-b)(a+b)$

$x^8-\frac{1}{16}$

$=(x^4)^2-\frac{1}{4^{-2}}$

$=(x^4)^2-(\frac{1}{4^{-1}})^2$

then applying the rule.

$=(x^4-\frac{1}{4})(x^4+\frac{1}{4})$

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