Write the value of `(1/2)^3+(1/3)^3-\ (5/6)^3`

Options

A. `5/12`
B. `(-5)/12`
C. `1/12`
D. `(-1)/12`

Answer

To solve the expression \((\frac{1}{2})^3 + (\frac{1}{3})^3 - (\frac{5}{6})^3\), we can utilize the identity for the sum of cubes:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 1: Identify \(a\), \(b\), and \(c\)
We can assign:
- \(a = \frac{1}{2}\)
- \(b = \frac{1}{3}\)
- \(c = -\frac{5}{6}\)

### Step 2: Calculate \(a + b + c\)
Now, we need to calculate \(a + b + c\):
\[ a + b + c = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 2, 3, and 6 is 6.

Converting each term:
\[ \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6}, \quad -\frac{5}{6} = -\frac{5}{6} \]

Now substituting:
\[ a + b + c = \frac{3}{6} + \frac{2}{6} - \frac{5}{6} = \frac{3 + 2 - 5}{6} = \frac{0}{6} = 0 \]

### Step 3: Calculate \(a^2 + b^2 + c^2\)
Next, we calculate \(a^2 + b^2 + c^2\):
\[ a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad b^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}, \quad c^2 = \left(-\frac{5}{6}\right)^2 = \frac{25}{36} \]

Finding a common denominator for these fractions (the LCM of 4, 9, and 36 is 36):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36}, \quad \frac{25}{36} = \frac{25}{36} \]

Adding these:
\[ a^2 + b^2 + c^2 = \frac{9}{36} + \frac{4}{36} + \frac{25}{36} = \frac{9 + 4 + 25}{36} = \frac{38}{36} = \frac{19}{18} \]

### Step 4: Calculate \(ab + ac + bc\)
Now we calculate \(ab + ac + bc\):
\[ ab = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}, \quad ac = \frac{1}{2} \cdot -\frac{5}{6} = -\frac{5}{12}, \quad bc = \frac{1}{3} \cdot -\frac{5}{6} = -\frac{5}{18} \]

Finding a common denominator for these fractions (the LCM of 6, 12, and 18 is 36):
\[ \frac{1}{6} = \frac{6}{36}, \quad -\frac{5}{12} = -\frac{15}{36}, \quad -\frac{5}{18} = -\frac{10}{36} \]

Adding these:
\[ ab + ac + bc = \frac{6}{36} - \frac{15}{36} - \frac{10}{36} = \frac{6 - 15 - 10}{36} = \frac{-19}{36} \]

### Step 5: Substitute into the identity
Now we substitute back into the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Since \(a + b + c = 0\), we have:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]
Thus:
\[ a^3 + b^3 + c^3 = 3abc \]

### Step 6: Calculate \(3abc\)
Now we calculate \(3abc\):
\[ abc = \frac{1}{2} \cdot \frac{1}{3} \cdot -\frac{5}{6} = -\frac{5}{36} \]
So:
\[ 3abc = 3 \cdot -\frac{5}{36} = -\frac{15}{36} = -\frac{5}{12} \]

### Final Answer
Thus, the value of the expression \((\frac{1}{2})^3 + (\frac{1}{3})^3 - (\frac{5}{6})^3\) is:
\[ \boxed{-\frac{5}{12}} \]