Question 175719
I'm assuming you're asked to factor the expression.
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Starting with:
8X^3y^6+27
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Notice, I can rewrite the above as:
(2xy^2)^3+(3)^3
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Now, it is a "sum of cubes" -- a special factor... where:
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
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In your case,
a = 2xy^2
b = 3
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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)