How i can factor #8*(a^(6n))+27*(b^(3m))# ?
1 Answer
Sep 9, 2015
Explanation:
Here's what I'd try.
Notice that you can write
#8 = 2""^3" "# and#" "27 = 3""^3#
which means that you can rewrite the original expression as
#2^3 * a^(6n) + 3^3 * b^(3m)#
You can do the same for
#a^(6n) = (a^(2n))^3" "# and#" "b^(3m) = (b^m)^3#
This will give you
#2^3 * (a^(2n))^3 + 3^3 * (b^m)^3 = (2 * a^(2n))^3 + (3 * b^m)^3#
Now you can use the sum of cubes factoring formula
#color(blue)(a^3 + b^3 = (a+b) * (a^2 - ab + b^3))#
to get
#(2a^(2n) + 3b^m) * [(2a^(2n))^2 - 2a^(2n) * 3b^(m) + (3b^(m))^2]#
#(2a^(2n) + 3b^(m)) * (4a^(4n)-6a^(2n)b^m + 9b^(2m))#
