If `x^(1//3)+ y ^(1//3)+z^(1//3)= 0 " then what is "(x+y+ z)^(3)` equal to ?

Options

A. 1
B. 3
C. 3xyz
D. 27 xyz

Explanation

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

**Step 1:** Start with the equation:
\[ x^{1/3} + y^{1/3} + z^{1/3} = 0 \]

**Step 2:** Rearrange the equation to isolate one of the cube roots:
\[ x^{1/3} = - (y^{1/3} + z^{1/3}) \]

**Step 3:** Now, we will cube both sides of the equation. Recall that \((a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)\). In our case, let \(a = x^{1/3}\), \(b = y^{1/3}\), and \(c = z^{1/3}\):
\[ (x^{1/3})^3 = (- (y^{1/3} + z^{1/3}))^3 \]

**Step 4:** Simplifying the left side gives:
\[ x = - (y^{1/3} + z^{1/3})^3 \]

**Step 5:** Now, we can expand the right side using the binomial theorem:
\[ (- (y^{1/3} + z^{1/3}))^3 = - (y^{1/3})^3 - 3(y^{1/3})^2(z^{1/3}) - 3(y^{1/3})(z^{1/3})^2 - (z^{1/3})^3 \]
This simplifies to:
\[ = - (y + z + 3y^{1/3}z^{1/3}(y^{1/3} + z^{1/3})) \]

**Step 6:** Now, we can express \(x + y + z\) in terms of \(y\) and \(z\):
\[ x + y + z = - (y + z + 3y^{1/3}z^{1/3}(y^{1/3} + z^{1/3})) + y + z \]

**Step 7:** Since \(x^{1/3} + y^{1/3} + z^{1/3} = 0\), we can conclude that:
\[ x + y + z = 0 \]

**Step 8:** Finally, we need to find \((x + y + z)^3\):
\[ (x + y + z)^3 = 0^3 = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]