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

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

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Answer 1 · 2016-02-05T00:23:34

$27 + 8 x^{3} = (3 + 2 x) (9 - 6 x + 4 x^{2}) = (2 x + 3) (4 x^{2} - 6 x + 9)$ $27+8x^3=(3+2x)(9-6x+4x^2)=(2x+3)(4x^2-6x+9)$

Explanation:

The sum of two cubes can be factored as:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$

For the problem at hand, $27 + 8 x^{3} = 3^{3} + {(2 x)}^{3}$ $27+8x^3=3^3+(2x)^3$ so that $a = 3$ $a=3$ and $b = 2 x$ $b=2x$ . Therefore

$27 + 8 x^{3} = (3 + 2 x) (9 - 6 x + 4 x^{2}) = (2 x + 3) (4 x^{2} - 6 x + 9)$ $27+8x^3=(3+2x)(9-6x+4x^2)=(2x+3)(4x^2-6x+9)$

Answer 2 · 2016-02-05T00:24:39

It is $27 + 8 x^{3} = (3 + 2 x) \cdot (9 - 6 x + 4 x^{2})$ $27+8x^3= (3+2x)*(9-6x+4x^2)$

Explanation:

Rewrite this as follows

$27 + 8 x^{3} = (3^{3}) + {(2 x)}^{3} = (3 + 2 x) \cdot (3^{2} - 6 x + 4 x^{2}) = (3 + 2 x) \cdot (9 - 6 x + 4 x^{2})$ $27+8x^3=(3^3)+(2x)^3=(3+2x)*(3^2-6x+4x^2)= (3+2x)*(9-6x+4x^2)$

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

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