If ` a^(4) + b^(4) = a^(2) b^(2) , ` then ` (a^(6) + b^(6))` equals

Options

A. 0
B. 1
C. `a^(2) + b^(2)`
D. `a^(2) b^(4) + a^(4) b^(2)`

Answer

To solve the equation \( a^{4} + b^{4} = a^{2} b^{2} \) and find \( a^{6} + b^{6} \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ a^{4} + b^{4} = a^{2} b^{2} \]

### Step 2: Rewrite \( a^{6} + b^{6} \)
We can express \( a^{6} + b^{6} \) in terms of \( a^{2} \) and \( b^{2} \):
\[ a^{6} + b^{6} = (a^{2})^{3} + (b^{2})^{3} \]
This can be factored using the sum of cubes formula:
\[ x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2}) \]
where \( x = a^{2} \) and \( y = b^{2} \).

### Step 3: Apply the sum of cubes formula
Thus, we have:
\[ a^{6} + b^{6} = (a^{2} + b^{2})((a^{2})^{2} - a^{2}b^{2} + (b^{2})^{2}) = (a^{2} + b^{2})(a^{4} - a^{2}b^{2} + b^{4}) \]

### Step 4: Substitute \( a^{4} + b^{4} \)
From the original equation, we know:
\[ a^{4} + b^{4} = a^{2}b^{2} \]
Thus, we can substitute this into our expression:
\[ a^{4} - a^{2}b^{2} + b^{4} = (a^{4} + b^{4}) - a^{2}b^{2} = a^{2}b^{2} - a^{2}b^{2} = 0 \]

### Step 5: Substitute back into the equation
Now substituting back, we have:
\[ a^{6} + b^{6} = (a^{2} + b^{2})(0) = 0 \]

### Conclusion
Therefore, the value of \( a^{6} + b^{6} \) is:
\[ \boxed{0} \]