# Difference of Squares

• Aug 25th 2008, 08:23 AM
cmf0106
Difference of Squares
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
• Aug 25th 2008, 08:26 AM
o_O
Note that $\displaystyle 2^6 = 64$ and the property of powers: $\displaystyle \frac{a^{n}}{b^{n}} = \left(\frac{a}{b}\right)^n$

So: $\displaystyle \frac{1}{64} = \frac{1^6}{2^6} = \left(\frac{1}{2}\right)^6$
• Aug 25th 2008, 08:27 AM
Jhevon
Quote:

Originally Posted by 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

no big deal here. using the rules to distribute the exponent through a rational function, $\displaystyle \left( \frac 12 \right)^6 = \frac {1^6}{2^6} = \frac 1{64}$ .....since $\displaystyle \left( \frac xy \right)^a = \frac {x^a}{y^a}$
• Aug 25th 2008, 08:31 AM
cmf0106
Thank you very much guys. Honestly, in the whole exponent chapter this rule was never stated believe it or not (Headbang)

Also once we established $\displaystyle x^6-(\frac{1}{2})^6$ why is it equal to $\displaystyle (x^3-\frac{1}{2})(x^3+\frac{1}{2})$ shouldnt the 6 exponent be divided by 2 resulting in $\displaystyle (x^3-\frac{1}{2}^3)(x^3+\frac{1}{2}^3)$
• Aug 25th 2008, 08:45 AM
masters
Quote:

Originally Posted by cmf0106
Thank you very much guys. Honestly, in the whole exponent chapter this rule was never stated believe it or not (Headbang)

Also once we established $\displaystyle x^6-(\frac{1}{2})^6$ why is it equal to $\displaystyle (x^3-\frac{1}{2})(x^3+\frac{1}{2})$ shouldnt the 6 exponent be divided by 2 resulting in $\displaystyle (x^3-\frac{1}{2}^3)(x^3+\frac{1}{2}^3)$

$\displaystyle (x^3-\frac{1}{8})(x^3+\frac{1}{8})$

Now you have two factors. The first factor is the difference of two cubes and the second factor is the sum of two cubes. Each of these factors can be factored further.
• Aug 25th 2008, 08:54 AM
cmf0106
so irrespective of the fact the book did not factor completely would $\displaystyle (x^3-\frac{1}{2})(x^3-\frac{1}{2})$ be correct? If so how do you arrive at that particular answer

Also could someone walk me through this problem as well $\displaystyle 16x^4-1=(2x)^4-1^4=(4x^2-1)(4x^2+1)=(2x-1)(2x+1)(4x^2+1)$
Lost in this one as well unfortunately (Crying). I understand breaking down $\displaystyle 16x^4-1=(2x)^4-1^4$ but how that yields $\displaystyle (4x^2-1)(4x^2+1)$ I am not certain.
• Aug 25th 2008, 09:06 AM
masters
Quote:

Originally Posted by cmf0106
so irrespective of the fact the book did not factor completely would $\displaystyle (x^3-\frac{1}{{\color{red}8}})(x^3{\color{red}+}\frac{1 }{{\color{red}8}})$ be correct? If so how do you arrive at that particular answer? To continue, you would have to know the model for the difference and sum of cubes. Is that part of your lesson?

Also could someone walk me through this problem as well $\displaystyle 16x^4-1=(2x)^4-1^4=(4x^2-1)(4x^2+1)=(2x-1)(2x+1)(4x^2+1)$
Lost in this one as well unfortunately (Crying). I understand breaking down $\displaystyle 16x^4-1=(2x)^4-1^4$ but how that yields $\displaystyle (4x^2-1)(4x^2+1)$ I am not certain.

Maybe you could look at it this way:

$\displaystyle 16x^4-1=(4x^2)^2-(1^2)^2$ which now looks like the difference between two squares.
• Aug 25th 2008, 09:16 AM
cmf0106
Quote:

Originally Posted by masters
Maybe you could look at it this way:

$\displaystyle 16x^4-1=(4x^2)^2-(1^2)^2$ which now looks like the difference between two squares.

With all due respect Masters, im afraid that does not help. Its most likely because of my own ignorance, but this book does tend to skip around alot.
• Aug 25th 2008, 09:24 AM
masters
Quote:

Originally Posted by cmf0106
With all due respect Masters, im afraid that does not help. Its most likely because of my own ignorance, but this book does tend to skip around alot.

$\displaystyle 16x^4-1=(4x^2)^2-(1^2)^2$

Let $\displaystyle u=4x^2$ and $\displaystyle v=1^2$

$\displaystyle u^2-v^2=(u-v)(u+v)$

Now substutute back in for u and v:

$\displaystyle (4x^2-1)(4x^2+1)$

Remember 1 raised to any power is 1.
• Aug 25th 2008, 09:28 AM
masters
This is a handy site for understanding the factoring of the difference of two squares and also has links for the sum and difference of cubes as well:

Special Factoring: Differences of Squares