`x^6-729`

Answer

To factor the expression \( x^6 - 729 \), we can follow these steps:

### Step 1: Recognize the difference of cubes
The expression \( x^6 - 729 \) can be rewritten as a difference of cubes. We note that \( 729 = 9^3 = (3^2)^3 = 3^6 \). Thus, we can express the equation as:

\[ x^6 - 729 = x^6 - (3^2)^3 \]

### Step 2: Apply the difference of cubes formula
The difference of cubes formula states that \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Here, we can let \( a = x^2 \) and \( b = 3 \):

\[ x^6 - 729 = (x^2 - 3)(x^4 + 3x^2 + 9) \]

### Step 3: Factor further if possible
Now, we need to check if \( x^4 + 3x^2 + 9 \) can be factored further. We can use the quadratic formula to check if it has real roots:

The expression \( x^4 + 3x^2 + 9 \) can be treated as a quadratic in terms of \( y = x^2 \):

\[ y^2 + 3y + 9 \]

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 36}}{2} = \frac{-3 \pm \sqrt{-27}}{2} \]

Since the discriminant is negative (\(-27\)), \( x^4 + 3x^2 + 9 \) does not have real roots and cannot be factored further over the real numbers.

### Final Answer
Thus, the complete factorization of \( x^6 - 729 \) is:

\[ x^6 - 729 = (x^2 - 3)(x^4 + 3x^2 + 9) \]

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