# Algebraic expressions

• Mar 20th 2011, 07:22 AM
ikislav
Algebraic expressions
Hello everyone.

I got the following problem:

If a *b != 0 and a != b, then the following expression:

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

equals:
a) $\displaystyle a^2+ab+b^2$
b) $\displaystyle a-b$
c) $\displaystyle a+b$
d) $\displaystyle \frac{1}{a}-\frac{1}{b}$
e) $\displaystyle \frac{1}{a}+\frac{1}{b}$

So, I started like this:

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

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

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

and that's where I'm stuck :(
• Mar 20th 2011, 08:00 AM
skeeter
Quote:

Originally Posted by ikislav
Hello everyone.

I got the following problem:

If a *b != 0 and a != b, then the following expression:

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

equals:
a) $\displaystyle a^2+ab+b^2$
b) $\displaystyle a-b$
c) $\displaystyle a+b$
d) $\displaystyle \frac{1}{a}-\frac{1}{b}$
e) $\displaystyle \frac{1}{a}+\frac{1}{b}$

So, I started like this:

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

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

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

and that's where I'm stuck :(

I assume the following is the correct problem statement ...

If $\displaystyle a \cdot b \ne = 0$ and $\displaystyle a \ne b$, then the following expression:

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

equals:
a) $\displaystyle a^2+ab+b^2$
b) $\displaystyle a-b$
c) $\displaystyle a+b$
d) $\displaystyle \frac{1}{a}-\frac{1}{b}$
e) $\displaystyle \frac{1}{a}+\frac{1}{b}$

...

$\displaystyle \left[\frac{a^2 - 2ab + b^2}{ab}+3\right] \cdot \frac{a^2-b^2}{a^3-b^3}$

$\displaystyle \left[\frac{a^2 - 2ab + b^2}{ab}+\frac{3ab}{ab}\right] \cdot \frac{(a-b)(a+b)}{(a-b)(a^2+ab+b^2)}$

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

can you finish?
• Mar 20th 2011, 08:03 AM
Plato
Quote:

Originally Posted by ikislav
If a *b != 0 and a != b, then the following expression: $\displaystyle (\frac{(a-b)^2}{ab}+3)*(\frac{a}{b}-\frac{b}{a}):\frac{a^3-b^3}{ab}$

Note two things: $\displaystyle \left( {\dfrac{{a^2 - 2ab + b^2 }}{{ab}} + 3} \right) = \left( {\dfrac{{a^2 + ab + b^2 }}{{ab}}} \right)$

$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)~.$
• Mar 20th 2011, 08:21 AM
ikislav
Quote:

Originally Posted by skeeter
I assume the following is the correct problem statement ...

If $\displaystyle a \cdot b \ne = 0$ and $\displaystyle a \ne b$, then the following expression:

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

equals:
a) $\displaystyle a^2+ab+b^2$
b) $\displaystyle a-b$
c) $\displaystyle a+b$
d) $\displaystyle \frac{1}{a}-\frac{1}{b}$
e) $\displaystyle \frac{1}{a}+\frac{1}{b}$

...

$\displaystyle \left[\frac{a^2 - 2ab + b^2}{ab}+3\right] \cdot \frac{a^2-b^2}{a^3-b^3}$

$\displaystyle \left[\frac{a^2 - 2ab + b^2}{ab}+\frac{3ab}{ab}\right] \cdot \frac{(a-b)(a+b)}{(a-b)(a^2+ab+b^2)}$

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

can you finish?

Ok, so than that's:

$\displaystyle \frac{a+b}{ab}$

Is that right? But it is not among the given choice of answers.
• Mar 20th 2011, 08:27 AM
skeeter
Quote:

Originally Posted by ikislav
Ok, so than that's:

$\displaystyle \frac{a+b}{ab}$

Is that right? But it is not among the given choice of answers.

yes, it is ...
• Mar 20th 2011, 08:39 AM
ikislav
Quote:

Originally Posted by skeeter
yes, it is ...

Ok, if I multiply with $\displaystyle ab$, I get $\displaystyle a^2b+ab^2$. I am at a loss here.
• Mar 20th 2011, 08:41 AM
skeeter
$\displaystyle \dfrac{a+b}{ab} = \dfrac{a}{ab} + \dfrac{b}{ab} =$ ... one more step
• Mar 20th 2011, 09:27 AM
ikislav
Quote:

Originally Posted by skeeter
$\displaystyle \dfrac{a+b}{ab} = \dfrac{a}{ab} + \dfrac{b}{ab} =$ ... one more step

I got it:

$\displaystyle \dfrac{a+b}{ab} = \dfrac{a}{ab} + \dfrac{b}{ab} = \dfrac{1}{b} + \dfrac{1}{a}$

Right answer is the one under "e" :)