How do you factor completely $27 x^{3} + 8$ $27x^3 + 8$ ?

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Answer 1 · 2018-03-25T05:59:10

$(3 x + 2) (9 x^{2} - 6 x + 4)$ $(3x+2)(9x^2-6x+4)$

Before we start, notice this: $27 x^{3}$ $27x^3$ and $8$ $8$ are both perfect cubes of $3 x$ $3x$ and $2$ $2$ respectively.

$27 x^{3} + 8$ $27x^3+8$

Now, apply sum of cubes rule: $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$ ,

$(3 x + 2) (9 x^{2} - 6 x + 4)$ $(3x+2)(9x^2-6x+4)$

How do you factor completely 27x3+827x^3 + 8?