SOLUTION: 8X^3y^6+27

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Question 175719: 8X^3y^6+27
Found 2 solutions by EMStelley, nerdybill:
Answer by EMStelley(208) About Me  (Show Source):
You can put this solution on YOUR website!
I believe what you are asking is to factor
8x%5E3y%5E6+%2B+27
Notice that each term can be written using cubes.
8x%5E3y%5E6+%2B+27+=+%282%5E3%29%28x%5E3%29%28y%5E2%29%5E3+%2B+3%5E3
So we have that
8x%5E3y%5E6+%2B+27+=+%282xy%5E2%29%5E3+%2B+3%5E3
Now, the sum of cubes can be factored as following:
a%5E3%2Bb%5E3=%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29
So, here a=2xy%5E2 and b=3

Simplified, this is
%282xy%5E2+%2B+3%29%284x%5E2y%5E4+-+6xy%5E2+%2B+9%29

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming you're asked to factor the expression.
.
Starting with:
8X^3y^6+27
.
Notice, I can rewrite the above as:
(2xy^2)^3+(3)^3
.
Now, it is a "sum of cubes" -- a special factor... where:
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
.
In your case,
a = 2xy^2
b = 3
.
Plugging the above into:
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
we get:
(2xy^2)^3 + 3^3 = (2xy^2 + 3)((2xy^2)^2 – 3(2xy^2) + 3^2)
(2xy^2)^3 + 3^3 = (2xy^2 + 3)(4x^2y^4 – 6xy^2 + 9)