Math Help - Difference of Squares

1. Difference of Squares

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

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

$x^6-\frac{1}{64}=x^6-(\frac{1}{2})^6=(x^3-\frac{1}{2})(x^3+\frac{1}{2})$
If someone could walk me through this it would be greatly appreciated,
Many Thanks

2. Note that $2^6 = 64$ and the property of powers: $\frac{a^{n}}{b^{n}} = \left(\frac{a}{b}\right)^n$

So: $\frac{1}{64} = \frac{1^6}{2^6} = \left(\frac{1}{2}\right)^6$

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

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

$x^6-\frac{1}{64}=x^6-(\frac{1}{2})^6=(x^3-\frac{1}{2})(x^3+\frac{1}{2})$
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, $\left( \frac 12 \right)^6 = \frac {1^6}{2^6} = \frac 1{64}$ .....since $\left( \frac xy \right)^a = \frac {x^a}{y^a}$

4. Thank you very much guys. Honestly, in the whole exponent chapter this rule was never stated believe it or not

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

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

Also once we established $x^6-(\frac{1}{2})^6$ why is it equal to $(x^3-\frac{1}{2})(x^3+\frac{1}{2})$ shouldnt the 6 exponent be divided by 2 resulting in $(x^3-\frac{1}{2}^3)(x^3+\frac{1}{2}^3)$
$(x^3-\frac{1}{8})(x^3+\frac{1}{8})$

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.

6. so irrespective of the fact the book did not factor completely would $(x^3-\frac{1}{2})(x^3-\frac{1}{2})$ be correct? If so how do you arrive at that particular answer

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

7. Originally Posted by cmf0106
so irrespective of the fact the book did not factor completely would $(x^3-\frac{1}{{\color{red}8}})(x^3{\color{red}+}\frac{1 }{{\color{red}8}})$ 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 $16x^4-1=(2x)^4-1^4=(4x^2-1)(4x^2+1)=(2x-1)(2x+1)(4x^2+1)$
Lost in this one as well unfortunately . I understand breaking down $16x^4-1=(2x)^4-1^4$ but how that yields $(4x^2-1)(4x^2+1)$ I am not certain.
Maybe you could look at it this way:

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

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

$16x^4-1=(4x^2)^2-(1^2)^2$ 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.

9. 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.
$16x^4-1=(4x^2)^2-(1^2)^2$

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

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

Now substutute back in for u and v:

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

Remember 1 raised to any power is 1.

10. 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