How do you simplify #7a ^ { 4} b \cdot ( - 3a ^ { 5} b ^ { 6} ) ^ { 2}#?

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Answer 1 · 2018-04-17T03:27:01

The answer is #63a^14b^13#.

Note: when the variables x, y, z, etc. are used, I am referring to a general rule that will work for every real value of x, y, z, etc.

First, you would expand out #(-3a^5b^6)^2#. Since #(x^y)^z=x^(y*z#, and #(v^xw^y)^z=v^(x*z)*w^(y*z)#, you can expan it as such:

#(-3)^2*a^(5*2)*b^(6*2)#, or #9a^10b^12# when simplified.

You can therefore substitute this into the original equation:

#7a^4b*9a^10b^12#

Since #x^y*x^z=x^(y+z)#, you can simplify this:

#(7*9)*a^(4+10)*b^(1+12)#, or, when simplified, #63a^14b^13#. This is your final answer.

Hope this helps!