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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

Respuesta :

[tex](3x^{-3})^2(-2xy)}[/tex]

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]
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