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

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

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Answer 1 · 2016-01-05T04:21:13

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

Explanation:

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

Since $8 u^{3}$ $8u^3$ and $27$ $27$ are cubes, we can rewrite the expression as ${(2 u)}^{3} + {(3)}^{3}$ $(2u)^3+(3)^3$ .

This a sum of two cubes with the form $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$ , where $a = 2 u$ $a=2u$ and $b = 3$ $b=3$ .

Substitute the values for $a$ $a$ and $b$ $b$ into the equation.

${(2 u)}^{3} + {(3)}^{3} = (2 u + 3) {(2 u)}^{2} - (2 u) (3) + {(3)}^{2}$ $(2u)^3+(3)^3=(2u+3)(2u)^2-(2u)(3)+(3)^2$

Simplify.

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

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

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