How do you add #(3s ^ { 3} + 45) + ( 6s ^ { 3} + s - 7)#?

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Answer 1 · 2017-04-06T01:45:41

#9s^3 + s + 38#

Given: #(3s^3+45)+(6s^3+s-7)#

To simplify, we need to remove the brackets first.

There is a #+# sign between the brackets, which means removing the second set of brackets will not change any signs inside.

#(3s^3+45)+(6s^3+s-7)=(3s^3+45)+6s^3+s-7#

and the same applies for the first set of brackets:

#(3s^3+45)+(6s^3+s-7)=3s^3+45+6s^3+s-7#

Now we can collect like terms, the add them:

#(3s^3+45)+(6s^3+s-7)=3s^3+6s^3+s+45-7#

#(3s^3+45)+(6s^3+s-7)=9s^3+s+38#