The polynomial, `x^(6)+64y^(6)` on factorization gives ___________.

Options

A. `(x^(2)+4y^(2))(x^(4)-4y^(4))`
B. `(x^(2)+4y^(2))(x^(4)-4x^(2)y^(2)+16y^(4))`
C. `(x^(2)-4y^(2))(x^(4)+4y^(4))`
D. `(x^(2)+4y^(2))(x^(4)+4x^(2)y^(2)+16y^(4))`

Answer

To factor the polynomial \( x^6 + 64y^6 \), we can follow these steps:

### Step 1: Recognize the form of the polynomial
The given polynomial is \( x^6 + 64y^6 \). We can see that it can be expressed as a sum of cubes since \( 64y^6 = (4y^2)^3 \).

### Step 2: Rewrite the polynomial
We can rewrite the polynomial as:
\[ x^6 + (4y^2)^3 \]
This allows us to identify \( a = x^2 \) and \( b = 4y^2 \).

### Step 3: Apply the sum of cubes formula
The sum of cubes can be factored using the identity:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Substituting \( a \) and \( b \):
\[ x^6 + 64y^6 = (x^2 + 4y^2)((x^2)^2 - (x^2)(4y^2) + (4y^2)^2) \]

### Step 4: Simplify the expression
Now we simplify the second factor:
\[ (x^2)^2 = x^4 \]
\[ -(x^2)(4y^2) = -4x^2y^2 \]
\[ (4y^2)^2 = 16y^4 \]
Putting it all together, we get:
\[ x^4 - 4x^2y^2 + 16y^4 \]

### Step 5: Write the final factorization
Thus, the complete factorization of the polynomial \( x^6 + 64y^6 \) is:
\[ (x^2 + 4y^2)(x^4 - 4x^2y^2 + 16y^4) \]

### Final Answer
The polynomial \( x^6 + 64y^6 \) on factorization gives:
\[ (x^2 + 4y^2)(x^4 - 4x^2y^2 + 16y^4) \]

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