Question

algebra
date feb 11
name diaz simon
factor cubes (show all work) (12.5 points each)
factor each (show all work).

1. $x^3 + 1 = 0$
2. $x^3 + 125 = 0$
3. $8x^3 - 27 = 0$
4. $27x^3 + 1 = 0$

-1-

Explanation:

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1) $x^3 + 1 = 0$

Step1: Recognize sum of cubes

$x^3 + 1^3 = 0$

Step2: Apply sum of cubes formula

Sum of cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$. Here $a=x, b=1$
$(x+1)(x^2 - x + 1) = 0$
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2) $x^3 + 125 = 0$

Step1: Recognize sum of cubes

$x^3 + 5^3 = 0$

Step2: Apply sum of cubes formula

Here $a=x, b=5$
$(x+5)(x^2 - 5x + 25) = 0$
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3) $8x^3 - 27 = 0$

Step1: Recognize difference of cubes

$(2x)^3 - 3^3 = 0$

Step2: Apply difference of cubes formula

Difference of cubes: $a^3-b^3=(a-b)(a^2+ab+b^2)$. Here $a=2x, b=3$
$(2x-3)(4x^2 + 6x + 9) = 0$
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4) $27x^3 + 1 = 0$

Step1: Recognize sum of cubes

$(3x)^3 + 1^3 = 0$

Step2: Apply sum of cubes formula

Here $a=3x, b=1$
$(3x+1)(9x^2 - 3x + 1) = 0$

Answer:

1. $(x+1)(x^2 - x + 1) = 0$
2. $(x+5)(x^2 - 5x + 25) = 0$
3. $(2x-3)(4x^2 + 6x + 9) = 0$
4. $(3x+1)(9x^2 - 3x + 1) = 0$