Question 1199464
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I'll do this one instead.  It's done exactly the same way as yours:

{{{(3x^3y)(5xy^5)-(12x^4y^2)(7y^4)}}}

Make sure that every letter shows an exponent.  If a letter doesn't
have an exponent showing, give it the exponent 1:

{{{(3x^3y^1)(5x^1y^5)-(12x^4y^2)(7y^4)}}}

For the first term:
3 times 5 is 15, {{{x^3}}} times {{{x^1}}} is {{{x^(3+1)}}} or {{{x^4}}}.
so the 1st term is {{{15x^4y^6}}}
For the second term (not using the - in front:
12 time 7 is 84, {{{y^2}}} times {{{y^4}}} is {{{y^(2+4)}}} or {{{y^6}}}.
so the 2nd term is {{{84x^4y^6}}}

{{{15x^4y^6 - 84x^4y^6}}}

Now since all the letters and exponents are the same in both those
terms, they are said to be LIKE TERMS, so we combine their coefficients
15-84 and get -69.

So the final answer is

{{{-69x^4y^6}}}

Now do yours the exact same way.

Edwin</pre>