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How do you factor: x^3 - 8 (x cubed minus 8)

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Lilithidn
[tex]x^3 - 8 =x^3-2^3 = (x-2) (x^2 + 2x + 2^2)=(x-2) (x^2 + 2x + 4)\\ \\ \\a^3 - b^3 = (a - b) (a^2 + ab + b^2)[/tex]


TheBlueFoxidn

Answer: (x - 2)(x² + 2x + 4)

Step-by-step explanation: In this problem, we're asked to factor x³ - 8.

Notice that x³ is a perfect cube and 8 is a perfect cube because 8 is 2 × 2 × 2 or 2³. So we have the difference of two cubes.

To factor the difference of two cubes, we use the following formula.

a³ - b³ can be factored as (a - b)(a² + ab + b²) and in this problem, since a³ is represented by x³, the value of a is x and since b³ is represented by 8, the value of b is 2.

So substituting x and 2 into the formula for a and b, we have

[(x) - (2)][(x)² + (x)(2) + (2)²] and notice that we changed the parentheses in the formula to brackets so that we're not dealing with too many sets of parentheses.

Next, simplifying inside the second set of brackets and changing the brackets back to parentheses, we have (x - 2)(x² + 2x + 4) which is our final answer.

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