Factories `8x^(3)+sqrt27 y^(3)`.

Answer

To factor the expression \( 8x^3 + \sqrt{27}y^3 \), we can follow these steps:

### Step 1: Rewrite the expression
First, we can simplify \( \sqrt{27} \):
\[ \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3} \]
Thus, the expression becomes:
\[ 8x^3 + 3\sqrt{3}y^3 \]

### Step 2: Identify \( a \) and \( b \)
Next, we can express \( 8x^3 \) and \( 3\sqrt{3}y^3 \) in the form of \( a^3 + b^3 \):
\[ a = 2x \quad \text{(since } (2x)^3 = 8x^3\text{)} \]
\[ b = \sqrt{3}y \quad \text{(since } (\sqrt{3}y)^3 = 3\sqrt{3}y^3\text{)} \]

### Step 3: Use the identity for \( a^3 + b^3 \)
We can use the algebraic identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Substituting \( a \) and \( b \) into the identity:
\[ 8x^3 + 3\sqrt{3}y^3 = (2x + \sqrt{3}y)((2x)^2 - (2x)(\sqrt{3}y) + (\sqrt{3}y)^2) \]

### Step 4: Simplify the expression
Now we calculate each part:
1. \( (2x)^2 = 4x^2 \)
2. \( (2x)(\sqrt{3}y) = 2\sqrt{3}xy \)
3. \( (\sqrt{3}y)^2 = 3y^2 \)

Putting it all together:
\[ = (2x + \sqrt{3}y)(4x^2 - 2\sqrt{3}xy + 3y^2) \]

### Final Answer
Thus, the factorization of \( 8x^3 + \sqrt{27}y^3 \) is:
\[ (2x + \sqrt{3}y)(4x^2 - 2\sqrt{3}xy + 3y^2) \]

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