# Math Help - Guidelines needed

1. ## Guidelines needed

The Problem:

$\frac {1}{a-b} - \frac {3ab}{a^3-b^3}$

What I have done:

$\frac {1}{a-b} - \frac {3ab}{a^3-b^3}$

$\frac {1}{a-b} - \frac {3ab}{(a-b)(a^2+ab+b^2)}$

I found the least common factor to be $(a-b)(a^2+ab+b^2)$

Then I multiplicated the first expression by $a^2+ab+b^2$ And therefore I have:

$\frac {1(a^2+ab+b^2)}{(a-b)(a^2+ab+b^2)} - \frac {3ab}{(a-b)(a^2+ab+b^2)}$

$\frac {a^2+ab+b^2-3ab}{(a-b)(a^2+ab+b^2)}$

And now I have:

$\frac {-3ab}{a-b}$

But the book answer is $\frac {a-b}{a^2+ab+b^2}$

Any guideline...???

2. Originally Posted by Alienis Back
The Problem:

$\frac {1}{a-b} - \frac {3ab}{a^3-b^3}$

What I have done:

$\frac {1}{a-b} - \frac {3ab}{a^3-b^3}$

$\frac {1}{a-b} - \frac {3ab}{(a-b)(a^2+ab+b^2)}$

I found the least common factor to be $(a-b)(a^2+ab+b^2)$

Then I multiplicated the first expression by $a^2+ab+b^2$ And therefore I have:

$\frac {1(a^2+ab+b^2)}{(a-b)(a^2+ab+b^2)} - \frac {3ab}{(a-b)(a^2+ab+b^2)}$

$\frac {a^2+ab+b^2-3ab}{(a-b)(a^2+ab+b^2)}$

And now I have:

$\frac {-3ab}{a-b}$

But the book answer is $\frac {a-b}{a^2+ab+b^2}$

Any guideline...???

Hi Alienis,

Let's pick it up from here:

$\frac {a^2+ab+b^2-3ab}{(a-b)(a^2+ab+b^2)}$

$\frac {a^2-2ab+b^2}{(a-b)(a^2+ab+b^2)}$

$\frac {(a-b)(a-b)}{(a-b)(a^2+ab+b^2)}$

The (a - b) cancels out and you have:

$\frac {a-b}{a^2+ab+b^2}$

3. You almost had it, just the last bit is wrong

simplify it like this
$\frac {a^2-2ab+b^2}{(a-b)(a^2+ab+b^2)}$
$\frac {(a-b)^2}{(a-b)(a^2+ab+b^2)}$
= $\frac {a-b}{a^2+ab+b^2}$

4. oh bugger, you just beat me to it

5. ## Don't get it yet...

And how did you turn $a^2+ab+b^2-3ab$ into $a^2-2ab+b^2$?

6. Originally Posted by Alienis Back
And how did you turn $a^2+ab+b^2-3ab$ into $a^2-2ab+b^2$?
$a^2+ab+b^2-3ab = a^2 + ab - 3ab + b^2 \, ....$