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What is the factored form of [tex]\(x^9 + 27\)[/tex]?

A. [tex]\((x^3 - 3)(x^6 + 3x^3 + 9)\)[/tex]

B. [tex]\((x^3 + 3)(x^6 - 3x^3 + 9)\)[/tex]

C. [tex]\((x - 3)^3 (x^6 + 3x^3 + 9)\)[/tex]

D. [tex]\((x + 3)^3 (x^6 - 3x^3 + 9)\)[/tex]

Sagot :

To find the factored form of the polynomial [tex]\(x^9 + 27\)[/tex], we can follow the steps of polynomial factorization with some known algebraic identities.

We start with the given polynomial:
[tex]\[ x^9 + 27 \][/tex]

Recognize that 27 is a perfect cube, i.e., [tex]\( 27 = 3^3 \)[/tex]. Therefore [tex]\( x^9 + 27 \)[/tex] can be written as:
[tex]\[ x^9 + 3^3 \][/tex]

We can use the sum of cubes formula which is:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]

In this case, consider [tex]\( x^9 \)[/tex] as [tex]\( (x^3)^3 \)[/tex]. Thus we can rewrite [tex]\( x^9 + 3^3 \)[/tex] as:
[tex]\[ (x^3)^3 + 3^3 \][/tex]

Applying the sum of cubes formula:
[tex]\[ (x^3 + 3)( (x^3)^2 - x^3 \cdot 3 + 3^2 ) \][/tex]

Simplifying the expressions inside the parentheses, we get:
[tex]\[ x^3 + 3 \][/tex]
[tex]\[ (x^3)^2 = x^6 \][/tex]
[tex]\[ -x^3 \cdot 3 = -3x^3 \][/tex]
[tex]\[ 3^2 = 9 \][/tex]

So, the factorization becomes:
[tex]\[ (x^3 + 3)(x^6 - 3x^3 + 9) \][/tex]

Now we can compare this result with the provided options:

A. [tex]\(\left(x^3 - 3\right)\left(x^6 + 3x^3 + 9\right)\)[/tex]
B. [tex]\(\left(x^3 + 3\right)\left(x^6 - 3x^3 + 9\right)\)[/tex]
C. [tex]\(\left(x - 3\right)^3\left(x^6 + 3x^3 + 9\right)\)[/tex]
D. [tex]\(\left(x + 3\right)^3\left(x^6 - 3x^3 + 9\right)\)[/tex]

The correct answer matches option B:
[tex]\[ (x^3 + 3)(x^6 - 3x^3 + 9) \][/tex]

Therefore, the answer to the question is:

B. [tex]\(\left(x^3 + 3\right)\left(x^6 - 3x^3 + 9\right)\)[/tex]
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