# Stuck on binomial expressions x.X

• Mar 29th 2013, 09:50 AM
Stuck on binomial expressions x.X
So I'm currently doing questions on binomial expressions, and I'm not sure as to how to do it after this >.> Any guidance? Thank you :]

This is the question: If $x - \frac{1}{x}= \frac{8}{3}$, find the value of $x^{3} - \frac{1}{x^{3}}$.
So this was what I tried to do:
I thought that it meant triple (we are currently doing expressions related to $\left ( a-b \right )^{3}= a^{3}-3a^{2}b + 3ab^{2}+b^{3}$ ^^)
So this was what I did:

$\left ( x-\frac{1}{x} \right )\left ( x-\frac{1}{x} \right )\left ( x-\frac{1}{x} \right )$

And then I simply substituted as given in the question:

$=\left ( \frac{8}{3} \right )\left ( \frac{8}{3} \right )\left ( \frac{8}{3} \right )$

And multiplied, and got the answer:
$=\frac{512}{27}= 26\frac{18}{27}$

But I'm pretty sure that what I have done is completely wrong.. any help?
• Mar 29th 2013, 10:17 AM
Plato
Re: Stuck on binomial expressions x.X
Quote:

Originally Posted by Lexadis
This is the question: If $x - \frac{1}{x}= \frac{8}{3}$, find the value of $x^{3} - \frac{1}{x^{3}}$.

\begin{align*} \left(x-\frac{1}{x}\right)^3&=x^3-3x+\frac{3}{x}-\frac{1}{x^3}\\&=\left(x^3-\frac{1}{x^3}\right)-3\left(x-\frac{1}{x}\right) \end{align*}
• Mar 29th 2013, 11:04 AM
Re: Stuck on binomial expressions x.X
Quote:

Originally Posted by Plato
\begin{align*} \left(x-\frac{1}{x}\right)^3&=x^3-3x+\frac{3}{x}-\frac{1}{x^3}\\&=\left(x^3-\frac{1}{x^3}\right)-3\left(x-\frac{1}{x}\right) \end{align*}

Oh I didn't think of it that way ;D
But then again, I'm confused as to how to proceed after that.
$= \left (x-\frac{1}{x} \right )^{3} - 3 * \frac{8}{3}$
$= x^{3}- \frac{1}{x^{3}} -8$
= ?
• Mar 29th 2013, 11:26 AM
MINOANMAN
Re: Stuck on binomial expressions x.X
it seems to me that you do not follow at all the instructions of Plato...
JUst substitute ...x^3-1/x^3=(8/3)^3+3(8/3) finally you will get 728/27 i.e 26.9629
• Mar 29th 2013, 12:03 PM
Re: Stuck on binomial expressions x.X
Oops sorry, didn't realize that x.X So, the final result is.. like this?

Quote:

Originally Posted by Plato
\begin{align*} \left(x-\frac{1}{x}\right)^3&=x^3-3x+\frac{3}{x}-\frac{1}{x^3}\\&=\left(x^3-\frac{1}{x^3}\right)-3\left(x-\frac{1}{x}\right) \end{align*}

......................................... $= \left ( \frac{8}{3} \right )^{3} - 3\left ( \frac{8}{3} \right )$
......................................... $= \frac{512}{27} - 8$
......................................... $= \frac{512 - 216}{27}$
......................................... $= \frac{296}{27}$
......................................... $= 10\frac{26}{27}$
• Mar 29th 2013, 12:13 PM
Plato
Re: Stuck on binomial expressions x.X
Quote:

Originally Posted by Lexadis
Oops sorry, didn't realize that x.X So, the final result is.. like this?

......................................... $= \left ( \frac{8}{3} \right )^{3} - 3\left ( \frac{8}{3} \right )$
......................................... $= \frac{512}{27} - 8$
......................................... $= \frac{512 - 216}{27}$
......................................... $= \frac{296}{27}$
......................................... $= 10\frac{26}{27}$

You missed a sign: $= \left ( \frac{8}{3} \right )^{3} {\color{red}+}~ 3\left ( \frac{8}{3} \right )$
• Mar 29th 2013, 12:19 PM
Re: Stuck on binomial expressions x.X
Quote:

Originally Posted by Plato
You missed a sign: $= \left ( \frac{8}{3} \right )^{3} {\color{red}+}~ 3\left ( \frac{8}{3} \right )$

I'm sorry but I just don't understand how a + comes in between.. any explanation?
• Mar 29th 2013, 01:15 PM
Plato
Re: Stuck on binomial expressions x.X
Quote:

Originally Posted by Lexadis
I'm sorry but I just don't understand how a + comes in between.. any explanation?

You have $\left(x-\frac{1}{x}\right)^3=\left(x^3-\frac{1}{x^3}\right)-3\left(x-\frac{1}{x}\right)$

that gives $\left(\frac{8}{3}\right)^3=\left(x^3-\frac{1}{x^3}\right){~\color{red}-}~3\left(\frac{8}{3}\right)$

Now solve for $\left(x^3-\frac{1}{x^3}\right)$.
• Mar 29th 2013, 01:37 PM
Re: Stuck on binomial expressions x.X
Quote:

Originally Posted by Plato
You have $\left(x-\frac{1}{x}\right)^3=\left(x^3-\frac{1}{x^3}\right)-3\left(x-\frac{1}{x}\right)$

that gives $\left(\frac{8}{3}\right)^3=\left(x^3-\frac{1}{x^3}\right){~\color{red}-}~3\left(\frac{8}{3}\right)$

Now solve for $\left(x^3-\frac{1}{x^3}\right)$.

Ooh I finally get it (so I think) so this was what I got:
................................... $\left(\frac{8}{3}\right)^3=\left(x^3-\frac{1}{x^3}\right){~\color{red}-}~3\left(\frac{8}{3}\right)$

................................... $\frac{512}{27}= x^{3}- \frac{1}{x^{3}}-3\left ( \frac{8}{3} \right )$

................................... $\frac{512}{27}= x^{3}- \frac{1}{x^{3}}-8$

................................... $\frac{512}{27}+8= x^{3}- \frac{1}{x^{3}}$

................................... $x^{3}- \frac{1}{x^{3}}= \frac{512+216}{27}$

................................... $x^{3}- \frac{1}{x^{3}}= \frac{728}{27}$

................................... $x^{3}- \frac{1}{x^{3}}= 26\frac{26}{27}$

Correct? :)