If `x ^(1//3) + y ^(1//3) = z ^(1//3)` then `( x + y - z) ^(3) + 27 xyz` is equal to

Options

A. 0
B. 1
C. `-1`
D. 27

Answer

To solve the problem, we start with the equation given:

\[ x^{1/3} + y^{1/3} = z^{1/3} \]

We need to find the value of the expression:

\[ (x + y - z)^3 + 27xyz \]

### Step 1: Rewrite the equation

Let \( a = x^{1/3} \), \( b = y^{1/3} \), and \( c = z^{1/3} \). Then, we can rewrite the equation as:

\[ a + b = c \]

### Step 2: Express \( x \), \( y \), and \( z \) in terms of \( a \), \( b \), and \( c \)

From the definitions of \( a \), \( b \), and \( c \), we have:

\[ x = a^3, \quad y = b^3, \quad z = c^3 \]

### Step 3: Substitute \( z \) in terms of \( a \) and \( b \)

Since \( c = a + b \), we can express \( z \) as:

\[ z = (a + b)^3 \]

### Step 4: Expand \( z \)

Using the binomial theorem, we expand \( z \):

\[ z = (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]

### Step 5: Substitute into the expression \( x + y - z \)

Now we substitute \( x \), \( y \), and \( z \) into the expression \( x + y - z \):

\[ x + y - z = a^3 + b^3 - (a^3 + 3a^2b + 3ab^2 + b^3) \]

This simplifies to:

\[ x + y - z = a^3 + b^3 - a^3 - b^3 - 3a^2b - 3ab^2 = -3a^2b - 3ab^2 = -3ab(a + b) \]

### Step 6: Cube the expression

Now we need to cube \( x + y - z \):

\[ (x + y - z)^3 = (-3ab(a + b))^3 = -27a^3b^3(a + b)^3 \]

### Step 7: Add \( 27xyz \)

Next, we calculate \( 27xyz \):

\[ 27xyz = 27(a^3)(b^3)(c^3) = 27a^3b^3(a + b)^3 \]

### Step 8: Combine the results

Now we combine the two parts:

\[ (x + y - z)^3 + 27xyz = -27a^3b^3(a + b)^3 + 27a^3b^3(a + b)^3 \]

This simplifies to:

\[ 0 \]

### Final Result

Thus, the final answer is:

\[ \boxed{0} \]