Question

select the correct answer.
what is the factored form of ( x^{9}+27 )?
( \bigcirc(x + 3)^{3}left(x^{6}-3 x^{3}+9
ight) )
( \bigcirc(x - 3)^{3}left(x^{6}+3 x^{3}+9
ight) )
( \bigcircleft(x^{3}+3
ight)left(x^{6}-3 x^{3}+9
ight) )
( \bigcircleft(x^{3}-3
ight)left(x^{6}+3 x^{3}+9
ight) )

Explanation:

Step1: Recall the sum of cubes formula

The sum of cubes formula is \(a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})\). Here, \(x^{9}+27=(x^{3})^{3}+3^{3}\). Let \(a=x^{3}\) and \(b = 3\).

Step2: Apply the sum of cubes formula

Using the formula \(a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})\), we substitute \(a=x^{3}\) and \(b = 3\). Then \((x^{3})^{3}+3^{3}=(x^{3}+3)[(x^{3})^{2}-x^{3}\times3 + 3^{2}]\).

Step3: Simplify the expression

Simplify \((x^{3}+3)[(x^{3})^{2}-x^{3}\times3 + 3^{2}]\) to get \((x^{3}+3)(x^{6}-3x^{3}+9)\).

Answer:

\((x^{3}+3)(x^{6}-3x^{3}+9)\) (the third option)