I couldn't solve for a

**Patrick_John** · June 28th 2007, 08:33 PM

When
$a - \frac{1}{a} = 2$
then
$a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} = ?$

The answer is $2$

I tried making a trinomial with $a - \frac{1}{a} = 2$ and I got $a^2 - 2a - 1$ , but after this I wasn't able to solve for a.

What should I do to get to the result $2$ ?

Thanks in advance.

**red_dog** · June 28th 2007, 09:11 PM

$\displaystyle a^3-2a^2-\frac{2}{a^2}-\frac{1}{a^3}=a^3-\frac{1}{a^3}-2\left(a^2+\frac{1}{a^2}\right)=$
$\displaystyle =\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right)-2\left(a^2+\frac{1}{a^2}\right)$ (1).
Now, if $\displaystyle a-\frac{1}{a}=2$ then $\displaystyle \left(a-\frac{1}{a}\right)^2=4$
$\displaystyle \Rightarrow a^2-2+\frac{1}{a^2}=4\Rightarrow a^2+\frac{1}{a^2}=4+2=6$ .
Replacing in (1) we have $2(6+1)-2\cdot 6=2$ .

**Jhevon** · June 28th 2007, 09:12 PM

Originally Posted by Patrick_John

When
$a - \frac{1}{a} = 2$
then
$a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} = ?$

The answer is $2$

I tried making a trinomial with $a - \frac{1}{a} = 2$ and I got $a^2 - 2a - 1$ , but after this I wasn't able to solve for a.

What should I do to get to the result $2$ ?

Thanks in advance.

If $a - \frac {1}{a} = 2$

Then: $a - 2 = \frac {1}{a}$ ..................(1)
and : $2 + \frac {1}{a} = a$ ....................(2)

Now: $a^3 - 2a^2 - \frac {2}{a^2} - \frac {1}{a^3} = a^2 (a - 2) - \frac {1}{a^2} \left( 2 + \frac {1}{a} \right)$ .............I factored out $a^2$ from the first two terms and $- \frac {1}{a^2}$ from the last two terms.

Now we can replace what's in brackets above with equations (1) and (2) respectively

$\Rightarrow a^3 - 2a^2 - \frac {2}{a^2} - \frac {1}{a^3} = a^2 \left( \frac {1}{a} \right) - \frac {1}{a^2} (a)$
................................ $= a - \frac {1}{a}$

................................ $= 2$

EDIT: Beaten by the great red_dog. Nice Solution, red_dog!

**Soroban** · June 28th 2007, 10:54 PM

Hello, Patrick_John!

When $a - \frac{1}{a} \:= \:2$ , then: . $a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} \:= \:?$

The answer is $2$

Square the equation: . $\left(a - \frac{1}{a}\right)^2 \:=\:2^2$

We have: . $a^2 - 2 + \frac{1}{a^2} \:=\:4\quad\Rightarrow\quad a^2 + \frac{1}{a^2} \:=\:6$

Multiply by $\text{-}2\!:\;\;\text{-}2a^2 - \frac{2}{a^2} \:=\:\text{-}12$ .[1]

Cube the equation: . $\left(a - \frac{1}{a}\right)^3 \:=\:2^3$

We have: . $a^3 - 3a + \frac{3}{a} - \frac{1}{a^3} \:=\:8\quad\Rightarrow\quad a^3 - 3\underbrace{\left(a - \frac{1}{a}\right)}_{\text{this is 2}} - \frac{1}{a^3} \:=\:8<br />$

. . . .Hence we have: . $a^3 - \frac{1}{a^3} \;=\;14$

. . . .Add [1]:. . . . . $\text{-}2a^2 - \frac{2}{a^2} \;=\;\text{-}12$

Therefore: . $a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} \;=\;2$

**Patrick_John** · June 29th 2007, 01:06 AM

Wow! Thanks a lot guys!

Can anybody recommend me a site or something where I can read more about this topic?

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