If `root3(x/729)+root3((8x)/729)+root3((27x)/5832)=1` , then find the valueof x.

Options

A. 1
B. 8
C. 3
D. 4

Answer

To solve the equation

\[ \sqrt[3]{\frac{x}{729}} + \sqrt[3]{\frac{8x}{729}} + \sqrt[3]{\frac{27x}{5832}} = 1, \]

we will follow these steps:

### Step 1: Rewrite the equation
First, we can express the cube roots in a more manageable form. We know that \(729 = 9^3\) and \(5832 = 18^3\). Thus, we rewrite the equation as:

\[ \sqrt[3]{\frac{x}{9^3}} + \sqrt[3]{\frac{8x}{9^3}} + \sqrt[3]{\frac{27x}{18^3}} = 1. \]

### Step 2: Factor out the common term
We can factor out \(\sqrt[3]{\frac{x}{9^3}}\) from the first two terms:

\[ \sqrt[3]{\frac{x}{9^3}} \left(1 + \sqrt[3]{8} + \sqrt[3]{\frac{27 \cdot 9^3}{18^3}}\right) = 1. \]

### Step 3: Simplify the cube roots
We know that \(\sqrt[3]{8} = 2\) and \(\sqrt[3]{27} = 3\). Now, we simplify \(\sqrt[3]{\frac{27 \cdot 729}{5832}}\):

\[ \sqrt[3]{\frac{27 \cdot 729}{5832}} = \sqrt[3]{\frac{27 \cdot 729}{18^3}} = \sqrt[3]{\frac{27 \cdot 729}{5832}} = \sqrt[3]{\frac{27 \cdot 729}{27 \cdot 216}} = \sqrt[3]{\frac{729}{216}} = \sqrt[3]{\frac{9^3}{6^3}} = \frac{9}{6} = \frac{3}{2}. \]

### Step 4: Substitute back into the equation
Now we substitute back into the equation:

\[ \sqrt[3]{\frac{x}{9^3}} \left(1 + 2 + \frac{3}{2}\right) = 1. \]

### Step 5: Combine the terms
Now we need to combine \(1 + 2 + \frac{3}{2}\):

\[ 1 + 2 = 3 \quad \text{and} \quad 3 + \frac{3}{2} = \frac{6}{2} + \frac{3}{2} = \frac{9}{2}. \]

So we have:

\[ \sqrt[3]{\frac{x}{729}} \cdot \frac{9}{2} = 1. \]

### Step 6: Solve for \(x\)
Now we can isolate \(\sqrt[3]{\frac{x}{729}}\):

\[ \sqrt[3]{\frac{x}{729}} = \frac{2}{9}. \]

Cubing both sides gives:

\[ \frac{x}{729} = \left(\frac{2}{9}\right)^3 = \frac{8}{729}. \]

### Step 7: Multiply both sides by 729
Now, multiply both sides by 729:

\[ x = 8. \]

### Final Answer
Thus, the value of \(x\) is

\[ \boxed{8}. \]