# Math Help - Need help simplifying and solving these...

1. ## Need help simplifying and solving these...

I need help with these.If possible please give me a step by step guide.Thanks

1. Simplify this: ((1-(((x^-n)+(y^-n))/((x^n)-(y^-n)))^-2) )
2. Calculate:x=((a^2)+(a^-2)) and y=((a^3)+(a^-3)) If a+(a^-1)=5 (a not 0).
3. (((((a^(-2x))-(a^-x)-6))/((a^(-2x))-4)) - (((a^(-x))-1)/(2-(a^(-x)))) - 2)

2. ## Re: Need help simplifying and solving these...

Originally Posted by loomi
I need help with these.If possible please give me a step by step guide.Thanks

1. Simplify this: ((1-(((x^-n)+(y^-n))/((x^n)-(y^-n)))^-2) )
2. Calculate:x=((a^2)+(a^-2)) and y=((a^3)+(a^-3)) If a+(a^-1)=5 (a not 0).
3. (((((a^(-2x))-(a^-x)-6))/((a^(-2x))-4)) - (((a^(-x))-1)/(2-(a^(-x)))) - 2)
Even though you have used brackets where they are needed, this is almost unreadable. A word of advice, if you have nested brackets (i.e. brackets inside brackets), the innermost brackets are round, the next set out are square, the next set out are curly, then back to round, etc...

4. ## Re: Need help simplifying and solving these...

Terribly presented.

Is the first one:
$(\frac{{1-x^{-n}+y^{-n}}}{(x^n-y^{-n})})^{-2}$

Or is it:

$1-(\frac{{x^{-n}+y^{-n}}}{(x^n-y^{-n})})^{-2}$

Or:

$1-(\frac{{x^{-n}+y^{-n}}}{(x^n-y^{-n})^{-2}})$

Or none of the above?

5. ## Re: Need help simplifying and solving these...

Never mind the first two here's the third one

7. ## Re: Need help simplifying and solving these...

1) $\frac{\1-(x^-n)+y^-n}{(x^n)-(y^-n)^-2)} = 1 - (x^-2n) - y^-3n$ *using product and quotient law

2) $X=a^2+a^-2 y=a^3 + a^-3$

$Solve a+(a^-1) =5$

$\frac{a^2 +1}{a} = 5a$

$a^2 + 1 = 5a^2$
$4a^2 = 1$
$a^2 = \frac{1}{4}$

Therefore, $a = \pm \frac{1}{2}$

8. ## Re: Need help simplifying and solving these...

Originally Posted by loomi
Never mind the first two here's the third one
Do you want to simplify the expression?

9. ## Re: Need help simplifying and solving these...

Originally Posted by loomi
Never mind the first two here's the third one
Let $u = a^{-x}$. Thus your question becomes: $\dfrac{u^2 - u - 6}{u^2-4} - \dfrac{u-1}{2-u} - 2$

You can factor both the numerator and denominator the first term (perhaps something will cancel).

10. ## Re: Need help simplifying and solving these...

$a+\frac{1}{a}=5$

$(a+\frac{1}{a})^2=25$

Expand and see what you get.

Then, once you know what $(a^2+\frac{1}{a^2})$ is,

Expand: $(a^2+\frac{1}{a^2})(a+\frac{1}{a})$

11. ## Re: Need help simplifying and solving these...

Originally Posted by loomi
The second one: find x and y: and

if

$x=a^2+\frac{1}{a^2}=a^2+\frac{1}{a^2}+2-2=(a+\frac{1}{a})^2-2=(5)^2-2$

$y=a^3+\frac{1}{a^3}=(a+\frac{1}{a})(a^2+\frac{1}{a ^2}-1)=(a+\frac{1}{a})(x-1)=5(22)$

EDIT: Oh! Quacky has already replied to it.

12. ## Re: Need help simplifying and solving these...

Hello, loomi!

Here's #2 . . . basically what Quacky said.

$\text{2. If }a + a^{-1} \,=\,5\;(a \ne 0)$

. . . $\text{(a) Calculate: }a^2 + a^{-2}$

We have: . . $a + a^{-1} \:=\:5$

Square: . $\left(a + a^{-1}\right)^2 \:=\:5^2$

n . . . . $a^2 + 2 + a^{-2} \:=\:5$

. . . . . . . . $a^2 + a^{-2} \:=\:3$

$\text{(b) Calculate: }a^3 + a^{-3}$

We have: . . . . . $a + a^{-1} \;=\;5$

$\text{Cube: }\qquad\quad\;\, (a+a^{-1})^3 \;=\;5^3$

n . $a^3 + 3a + 3a^{-1} + a^{-3} \;=\;125$

. . $a^2 + 3\underbrace{\left(a + a^{-1}\right)}_{\text{This is 5}} + a^{-3} \;=\;125$

n . . . . . . $a^3 + 15 + a^{-3} \;=\;125$

. . . . . . . . . . . $a^3 + a^{-3} \;=\;110$

13. ## Re: Need help simplifying and solving these...

Originally Posted by Soroban
Hello, loomi!

Here's #2 . . . basically what Quacky said.

We have: . . $a + a^{-1} \:=\:5$

Square: . $\left(a + a^{-1}\right)^2 \:=\:5^2$

n . . . . $a^2 + 2 + a^{-2} \:=\:5$

. . . . . . . . $a^2 + a^{-2} \:=\:3$

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Thanks but I think it was
5^2-2=23

14. ## Re: Need help simplifying and solving these...

I agree with you - Soroban just made a slight typo.

15. ## Re: Need help simplifying and solving these...

This thread is a mess. Posts without quotes, too many questions asked and re-asked. Links that have to be clicked on to see the complete question etc. It's all over the shop. Good luck to anyone else trying to follow it.

All members, PLEASE do NOT reply to posts that are going to create this sort of thread confusion. Report the post and let a Moderator sort things out.