^{-3}(x^{-2})}{(x^2)^{-4})
When you raise a base with an exponent to an exponent you multiply the exponents (ie
). For instance ^{-4}=x^{2*(-4)}=x^{-8})
^{-3*2}(x^{-2})}{(x)^{-4*2})
^{-6}(x^{-2})}{(x)^{-8})
When you multiply like bases with exponents, you add the exponents (ie
)
^{-6+(-2)}}{(x)^{-8})
Multiply the numerator by adding the exponents
When you divide like bases with exponents, you subtract the exponents (ie
)
^{-8-(-8)})
^{-8+8)})
^{0}=1)
After all of that, we get an answer of 1. This means that for any value of x, I should get 1 everytime
Check:
Let x equal any number. Lets make x equal to 2
^{-3}(x^{-2})}{(x^2)^{-4})
Plug in x=2
^{-3}(\frac{1}{4})}{(4)^{-4})
}{\frac{1}{4^4})
}{\frac{1}{256})

These cancel
If you want more proof, let x equal any other number and you will get 1. Hope this makes sense.