How can I solve this?

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1 Answer
Jun 23, 2017

D. #(xy-1)/(xy)#

Explanation:

#(1-x^-3y^-3)/(x^-2y^-2+x^-1y^-1+1)#

When you have negative exponents, you need to factor out the term with the smallest exponent.

In the numerator, it's #x^-3y^-3#. In the denominator it's #x^-2y^-2#.

#((x^-3y^-3)(x^3y^3-1))/((x^-2y^-2)(1+xy+x^2y^2))#

If you look at the second factor in the numerator, you can identify it as the cubic trinomial #a^3-b^3#.

#a^3-b^3=(a-b)(a^2+ab+b^2)#

Therefore, you can use this formula to simplify the second factor of the numerator.

#((x^-3y^-3)(xy-1)(x^2y^2+xy+1))/((x^-2y^-2)(1+xy+x^2y^2))#

Although they are written in different orders, the two right most factors in the numerator and denominator are the same. You can now cancel them out.

#((x^-3y^-3)(xy-1)cancel(x^2y^2+xy+1))/((x^-2y^-2)cancel(1+xy+x^2y^2))#

#((x^-3y^-3)(xy-1))/(x^-2y^-2)#

Last, you have one more factor that can be canceled out. You can do this by simplifying the first factor in the numerator.

#((x^-1y^-1)(x^-2y^-2)(xy-1))/(x^-2y^-2)#

#((x^-1y^-1)cancel(x^-2y^-2)(xy-1))/cancel(x^-2y^-2)#

Now you are left with this:

#(x^-1y^-1)(xy-1)#

You can rewrite the first term into a fraction and then simplify.

#1/(xy)(xy-1)#

#(xy-1)/(xy)#

If done correctly, you should get D as the right answer!