# Thread: Figuring out how all these equal N[3]?

1. ## Figuring out how all these equal N[3]?

''[3]'' = power of 3

(n-2)[3]

(n-2)[2]x6

(n-2)[12]

8

Do you expand the brackets? and then what?

2. ## Re: Figuring out how all these equal N[3]?

Originally Posted by imalouski
''[3]'' = power of 3

(n-2)[3]

(n-2)[2]x6

(n-2)[12]

8
Pray tell what any of the above means!
It makes no sense.

3. ## Re: Figuring out how all these equal N[3]?

Sorry, but that's what my teacher told us to write down and calculate how they equal N[3] - I'm as confused as you then haha

4. ## Re: Figuring out how all these equal N[3]?

Originally Posted by imalouski
Sorry, but that's what my teacher told us to write down and calculate how they equal N[3] - I'm as confused as you then haha
Help from anyone is probability useless to you.

5. ## Re: Figuring out how all these equal N[3]?

Hello, imalouski!

Figuring out how all these equal $N^3.$
I think it means: Show that these are "somethng cubed".

$(n-2)^3$
This is the cube of $(n-2)$

$(n-2)^2x^6$
$(n-2)x^6 \;=\;\big[(n-2)x^2\big]^3$

This is the cube of $(n-2)x^2$

$(n-2)^{12}$
$(n-2)^{12} \;=\;\big[(n-2)^4\big]^3$

This is the cube of $(n-2)^4$

$8$
$8 \;=\;2^3$

This is the cube of $2.$

6. ## Re: Figuring out how all these equal N[3]?

Would it make any difference if i changed the question to, ''Calculate why it equals N[3]"?