If $\displaystyle$\displaystyle \sqrt[3]{a}+\sqrt[3]{b}+$\displaystyle \sqrt[3]{c}=0$ then the value of $\displaystyle \left($$\displaystyle \frac{a+b+c}{3}$\right)^3 is equal to

Let us define:
$$x =$$$\displaystyle \sqrt[3]{a}$, \quad y = $\displaystyle \sqrt[3]{b}$, \quad z = $\displaystyle \sqrt[3]{c}$
According to the problem, we are given:
$$x + y + z = 0$$
Recall the standard algebraic identity: if the sum of three variables is zero ($\displaystyle x + y + z = 0$), then the sum of their cubes is equal to three times their product:
$$x^3 + y^3 + z^3 = 3xyz$$
Substitute the definitions of $x, y, z$ back into this identity:
\displaystyle &#x27; in math mode at position 150: …rt[3]{c}\right)̲重<br>" style="color:#cc0000">\left($\displaystyle \sqrt[3]{a}$\right)^3 + \left($\displaystyle \sqrt[3]{b}$\right)^3 + \left($\displaystyle \sqrt[3]{c}$\right)^3 = 3 \left($\displaystyle \sqrt[3]{a}$ $\displaystyle \cdot$ $\displaystyle \sqrt[3]{b}$ $\displaystyle \cdot$ $\displaystyle \sqrt[3]{c}$\right)\displaystyle 重&lt;br&gt;</span>a + b + c = 3 $\displaystyle \sqrt[3]{abc}$<span class="katex-error" title="ParseError: KaTeX parse error: Can&#x27;t use function &#x27;' in math mode at position 47: …he equation by \displaystyle ̲3:<br>" style="color:#cc0000"><br>Now, divide both sides of the equation by $\displaystyle 3$:<br>\displaystyle \frac{a + b + c}{3} = \sqrt[3]{abc}$$<br>To find the final value, cube both sides of the equation:<br>$$\left($\displaystyle \frac{a + b + c}{3}$\right)^3 = \left($\displaystyle \sqrt[3]{abc}$\right)^3 = abc$$
This matches **Option A**.
Hence, **Option A** is the correct answer.