# Thread: Difference of Squares question

1. ## Difference of Squares question

I have troubles with any difference of squares expression that incorporates a fraction.

For instance in the expression $\displaystyle \frac{1}{64}x^6-1$
=$\displaystyle (\frac{1}{8}x^3-1)(\frac{1}{8}x^3+1)$

I understand the$\displaystyle x^3$ here comes from $\displaystyle x^{6/2}$, however I have no clue how we get $\displaystyle \frac{1}{8}$ from $\displaystyle \frac{1}{64}$. Would someone please enlighten me?

Thank you

2. Originally Posted by allyourbass2212
I have troubles with any difference of squares expression that incorporates a fraction.

For instance in the expression $\displaystyle \frac{1}{64}x^6-1$
=$\displaystyle (\frac{1}{8}x^3-1)(\frac{1}{8}x^3+1)$

I understand the$\displaystyle x^3$ here comes from $\displaystyle x^{6/2}$, however I have no clue how we get $\displaystyle \frac{1}{8}$ from $\displaystyle \frac{1}{64}$. Would someone please enlighten me?

Thank you
$\displaystyle \frac{1}{64} = (\frac{1}{8})^2$

3. Thanks that helps, but it also creates another source of confusion

if the original problem was $\displaystyle \frac{1}{64}x^6-1$

Why are we using

Where is the ^2 for the $\displaystyle 1/8$? And why are we not using ^3 since $\displaystyle ^{6/2}=3$?

4. Originally Posted by allyourbass2212
Thanks that helps, but it also creates another source of confusion

if the original problem was $\displaystyle \frac{1}{64}x^6-1$

Why are we using

Where is the ^2 for the $\displaystyle 1/8$? And why are we not using ^3 since $\displaystyle ^{6/2}=3$?
For this problem you have to know that $\displaystyle (\frac{1}{8})^2 = \frac{1^2}{8^2} = \frac{1}{64}$

We do not use the cube because we're taking the square root of the entire $\displaystyle \frac{1}{64}x^6$ which means both the 1/64 and the x^6 should be raised to the power of 1/2.

Imagine that the 64 is raised to the power of 1. Like you did with the x^6 we can divide the exponent by 2 to give 64^(1/2) = 8

Interestingly (although off topic) this expression is also the difference of two cubes

5. Originally Posted by allyourbass2212
Thanks that helps, but it also creates another source of confusion

if the original problem was $\displaystyle \frac{1}{64}x^6-1$

Why are we using

Where is the ^2 for the $\displaystyle 1/8$? And why are we not using ^3 since $\displaystyle ^{6/2}=3$?
You can look at it reverse wise so that you can see what's happening:

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

$\displaystyle =\frac{1}{8}x^3\cdot\frac{1}{8}x^3-\frac{1}{8}x^3+\frac{1}{8}x^3-1=$

$\displaystyle \left(\frac{1}{8}\cdot\frac{1}{8}\right)\cdot{x^3} \cdot{x^3}-1$

$\displaystyle =\left(\frac{1}{8}\right)^2\cdot(x^3)^2-1$

$\displaystyle =\frac{1}{64}x^6-1$

Now read this again, but start from the bottom and work your wat to the top.

6. Thanks guys, what I have been doing is just taking the number or fraction and raising it to ^(1/2) and it has worked for every problem except this one, so I guess this trick doesnt always work.

For instance
$\displaystyle \frac{16}{81}x^4-16=(\frac{4}{9}x^2-4)(\frac{4}{9}x^2+4)=(\frac{2}{3}x-2)(\frac{2}{3}x+2)(\frac{4}{9}x^2+4)$

Taking $\displaystyle \frac{16}{81}^{1/2}$ unfortunately does not yield either of the simplified fractions listed above, rather $\displaystyle \frac{16}{9}$

Edit Update: Well I guess I take that back. For some reason in my TI-83 if I just enter $\displaystyle \frac{16}{81}^{1/2}$ it will yield $\displaystyle \frac{16}{9}$

but if I do $\displaystyle \frac{16}{81} = .1975308642$, turn it back into the fraction $\displaystyle \frac{16}{81}$ and then raise that to $\displaystyle ^{1/2}$ it will give me the correct fraction $\displaystyle \frac{4}{9}$

Very strange indeed, anyone know why the calculator might be doing this?

7. Just enter it into your calculator like this "(16/81)^(1/2)".

8. Originally Posted by allyourbass2212
Thanks guys, what I have been doing is just taking the number or fraction and raising it to ^(1/2) and it has worked for every problem except this one, so I guess this trick doesnt always work.

For instance
$\displaystyle \frac{16}{81}x^4-16=(\frac{4}{9}x^2-4)(\frac{4}{9}x^2+4)=(\frac{2}{3}x-2)(\frac{2}{3}x+2)(\frac{4}{9}x^2+4)$

Taking $\displaystyle \frac{16}{81}^{1/2}$ unfortunately does not yield either of the simplified fractions listed above, rather $\displaystyle \frac{16}{9}$

Edit Update: Well I guess I take that back. For some reason in my TI-83 if I just enter $\displaystyle \frac{16}{81}^{1/2}$ it will yield $\displaystyle \frac{16}{9}$
Sounds like you didn't use parentheses when you inputted into the calculator, as cmf0106 suggests. You really have to be careful about things like that. If you typed something like

16/81^(1/2)

then only the 81 is going to be raised to the 1/2 power because of order of operations. If you want to raise the entire fraction to the 1/2 power you need parentheses:

(16/81)^(1/2)

01