How do you factor $8a^3 + 27b^3 + 2a + 3b$ ?

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Answer 1 · 2016-03-29T00:18:26

$8a^3+27b^3+2a+3b=(2a+3b)(4a^2-6ab+9b^2+1)$

From the given expression
$8a^3+27b^3+2a+3b$

$8a^3+27b^3+2a+3b$

$(8a^3+27b^3)+(2a+3b)$

The first two terms can be factored by sum of two cubes formula

$x^3+y^3=(x+y)(x^2-xy+y^2)$

so that $8a^3+27b^3=(2a+3b)(4a^2-6ab+9b^2)$

Let us continue

$(2a+3b)(4a^2-6ab+9b^2)+(2a+3b)$

factor out the common term (2a+3b)

$(2a+3b)(4a^2-6ab+9b^2+1)$

therefore

$8a^3+27b^3+2a+3b=(2a+3b)(4a^2-6ab+9b^2+1)$

God bless...I hope the explanation is useful.

How do you factor 8a^3 + 27b^3 + 2a + 3b8a^3 + 27b^3 + 2a + 3b?