''[3]'' = power of 3
(n-2)[3]
(n-2)[2]x6
(n-2)[12]
8
Do you expand the brackets? and then what?
Hello, imalouski!
I think it means: Show that these are "somethng cubed".Figuring out how all these equal $\displaystyle N^3.$
This is the cube of $\displaystyle (n-2)$$\displaystyle (n-2)^3$
$\displaystyle (n-2)x^6 \;=\;\big[(n-2)x^2\big]^3$$\displaystyle (n-2)^2x^6$
This is the cube of $\displaystyle (n-2)x^2$
$\displaystyle (n-2)^{12} \;=\;\big[(n-2)^4\big]^3$$\displaystyle (n-2)^{12}$
This is the cube of $\displaystyle (n-2)^4$
$\displaystyle 8 \;=\;2^3$$\displaystyle 8$
This is the cube of $\displaystyle 2.$