I'm helping my son with his algebra (attempting to anyway). We've got a problem that I can't quite understand... Simplify the equation (3x^-3)^2(-2xy). I
The square (²) can be distributed to each factor inside those parentheses. 3² = 3 × 3 = 9 as for the x^-3, we want to multiply our exponents together, leaving x^-6.
[tex](9x^{-6})(-2xy)[/tex]
Note that since multiplication is commutative/associative, we can multiply these together in whatever order to simplify.
The x on the right can be rewritten as x¹. Multiply that and x^-6 together to get x^-5. (we just add the exponents) As for the 9 and -2, multiply those together to get -18.
Now we have [tex]-18x^{-5}y[/tex].
Lastly, if you don't want a negative exponent in your answer: It can be flipped to the bottom because of this property:
[tex]x^{-n}=\frac{1}{x^n}[/tex]
This leaves us with [tex]\boxed{\frac{-18y}{x^5}}[/tex]