How i can factor #8*(a^(6n))+27*(b^(3m))# ?

1 Answer
Sep 9, 2015

Answer:

#(2a^(2n) + 3b^(m)) * (4a^(4n)-6a^(2n)b^m + 9b^(2m))#

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))#