Question 130954
First let's simplify the given monomials individually


{{{(a^2)^6}}} Start with the first expression. 



{{{a^(2*6)}}} Multiply the outer exponent with the inner exponent. Remember {{{(x^y)^z=x^(y*z)}}}



{{{a^12}}} Now multiply the exponents



So {{{(a^2)^6}}}  simplifies to {{{a^12}}}.


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{{{((-2)^1a^4)^3}}} Start with the second expression. Note: {{{(-2a^4)^3}}} really looks like {{{((-2)^1a^4)^3}}}



{{{(-2)^(1*3)a^(4*3)}}} Now distribute the outer exponent 3 to each exponent in the parenthesis. Remember {{{(x^y)^z=x^(y*z)}}}



{{{(-2)^3a^12}}} Now multiply the exponents



{{{-8a^12}}} Now raise -2 to the 3rd power to get -8


So {{{(-2a^4)^3}}} simplifies to {{{-8a^12}}}.




a)



Now let's find the sum. To do that, simply add the two simplified expressions together



{{{a^12+(-8a^12)}}}



{{{-7a^12}}} Combine like terms



So  {{{a^12}}} and {{{-8a^12}}} add to {{{-7a^12}}}. In other words, {{{a^12+(-8a^12)=-7a^12}}}





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b)



Now let's find the product. To do that, simply multiply the two simplified expressions together



{{{(a^12)(-8a^12)}}}



{{{-8a^(12+12)}}} Add the exponents



{{{-8a^24}}} Simplify



So  {{{a^12}}} and {{{-8a^12}}} multiply to {{{-8a^24}}}. In other words, {{{(a^12)(-8a^12)=-8a^24}}}