• Mar 12th 2010, 04:21 PM
ejanderson
I've worked and re-worked this problem over the last couple of days and cannot understand where I'm going wrong. This is also the first time I'm attempting to use LaTeX so forgive me if I do it wrong.

Change the following in to common fractions:

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

Right away I determine the LCD = a-2b

So I then re-write the equation:

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

After doing the multiplication I get:

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

Combining like terms I get (I hope I've managed to keep all the information correct in the latex formatting):

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

The answer the book shows is:

$-\frac{a^2b+3ab^2-4b^3}{a-2b}$

Any help would be greatly appreciated and much thanked in advance as always. Thanks!

E
• Mar 12th 2010, 04:25 PM
TKHunny
Quote:

Originally Posted by ejanderson
After doing the multiplication I get:

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

So close. Check the very last sign in the numerator. Be careful!
• Mar 12th 2010, 07:35 PM
ejanderson

Since the last fraction $-\frac{a^3-2b^3}{a-2b}$ already contains the LCD, I don't do a thing to the numerator which should keep the $-a^3-2b^3$ the same... $-a-2b^3$.

Now, looking back to the third fraction $-\frac{b^2(a-2b)}{a-2b}$ that becomes: $-\frac{ab^2+2b^3}{a-2b}$

The part of the answer I'm missing is the $-4b^3$ which should come from combining like terms from the last two fractions. However, in the way I'm doing it the $+2b^3$ in the third fraction gets cancelled out by the $-2b^3$ in the last fraction.

I "know" that my problem is coming from one of the last two fractions in the way I'm multiplying or combining terms but I just don't see it.
• Mar 12th 2010, 07:43 PM
Quacky
I would rewrite the last fraction as:

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

Which will clearly become:

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

Which, I think, should give the correct answer?
• Mar 12th 2010, 07:52 PM
sa-ri-ga-ma
-(a^2 - 2b^3) = -a^3 +2b^3
• Mar 12th 2010, 08:33 PM
TKHunny
Quote:

Originally Posted by ejanderson
I "know" that my problem is coming from one of the last two fractions in the way I'm multiplying or combining terms but I just don't see it.

Please learn the Distributive Property of Multiplication over Addition. There is some confusion over the dual identity of that subtraction. Write it the grade school way and you will see it.

a - b = a + (-b)

It's ugly, but that's the way it is.

You can also run a test.

$5 - \frac{7-3}{2}$

$\frac{10 - 7 - 3}{2}\;\;$ or $\;\;\frac{10 - 7 + 3}{2}$