If `x^(3) + y^(3) = 16` and `x + y = 4`, then the value of `x^(4) + y^(4)` is :

Options

A. 48
B. 32
C. 64
D. 30

Answer

To solve the problem, we need to find the value of \( x^4 + y^4 \) given the equations \( x^3 + y^3 = 16 \) and \( x + y = 4 \).

### Step 1: Use the identity for the sum of cubes
We can use the identity for the sum of cubes:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
Given \( x + y = 4 \), we can substitute this into the equation:
\[ x^3 + y^3 = 4(x^2 - xy + y^2) \]
We know from the problem that \( x^3 + y^3 = 16 \). Therefore, we can set up the equation:
\[ 16 = 4(x^2 - xy + y^2) \]

### Step 2: Simplify the equation
Now, divide both sides by 4:
\[ 4 = x^2 - xy + y^2 \]

### Step 3: Express \( x^2 + y^2 \) in terms of \( xy \)
We can use the identity:
\[ x^2 + y^2 = (x + y)^2 - 2xy \]
Substituting \( x + y = 4 \):
\[ x^2 + y^2 = 4^2 - 2xy = 16 - 2xy \]
Now, substituting this into our earlier equation:
\[ 4 = (16 - 2xy) - xy \]
This simplifies to:
\[ 4 = 16 - 3xy \]

### Step 4: Solve for \( xy \)
Rearranging gives:
\[ 3xy = 16 - 4 \]
\[ 3xy = 12 \]
\[ xy = 4 \]

### Step 5: Find \( x^2 + y^2 \)
Now we can substitute \( xy = 4 \) back into the equation for \( x^2 + y^2 \):
\[ x^2 + y^2 = 16 - 2(4) = 16 - 8 = 8 \]

### Step 6: Use the identity for the sum of fourth powers
Now we can find \( x^4 + y^4 \) using the identity:
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2 \]
Substituting the values we found:
\[ x^4 + y^4 = (8)^2 - 2(4)^2 \]
Calculating this gives:
\[ x^4 + y^4 = 64 - 2(16) = 64 - 32 = 32 \]

### Final Answer
Thus, the value of \( x^4 + y^4 \) is \( \boxed{32} \).

---