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

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Answer 1 · 2015-09-09T23:42:45

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

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

$(2a^(2n) + 3b^m) * [(2a^(2n))^2 - 2a^(2n) * 3b^(m) + (3b^(m))^2]$

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

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