Evaluate:
`((0. 43) ^(3) + (1. 47) ^(3) + (1.1) ^(3) - 3 xx 0. 43 xx 1 . 47 xx 1.1)/(( ( 0. 43) ^(2) + (1. 47) ^(2) + (1.1) ^(2) -0. 43 xx 1. 43)- 0. 43 xx 1.1 - 1.47 xx 1.1)`

Options

A. `1.90`
B. `2.87`
C. 3
D. `3.47`

Answer

To evaluate the expression

\[ \frac{(0.43)^3 + (1.47)^3 + (1.1)^3 - 3 \cdot 0.43 \cdot 1.47 \cdot 1.1}{(0.43)^2 + (1.47)^2 + (1.1)^2 - 0.43 \cdot 1.47 - 0.43 \cdot 1.1 - 1.47 \cdot 1.1} \]

we can use the identity for the sum of cubes and the sum of squares.

### Step 1: Identify \(a\), \(b\), and \(c\)
Let:
- \(a = 0.43\)
- \(b = 1.47\)
- \(c = 1.1\)

### Step 2: Apply the sum of cubes formula
The formula for the sum of cubes states:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 3: Calculate \(a + b + c\)
\[ a + b + c = 0.43 + 1.47 + 1.1 = 3.0 \]

### Step 4: Calculate \(a^2 + b^2 + c^2\)
\[ a^2 = (0.43)^2 = 0.1849 \]
\[ b^2 = (1.47)^2 = 2.1609 \]
\[ c^2 = (1.1)^2 = 1.21 \]
\[ a^2 + b^2 + c^2 = 0.1849 + 2.1609 + 1.21 = 3.5558 \]

### Step 5: Calculate \(ab\), \(ac\), and \(bc\)
\[ ab = 0.43 \cdot 1.47 = 0.6321 \]
\[ ac = 0.43 \cdot 1.1 = 0.473 \]
\[ bc = 1.47 \cdot 1.1 = 1.617 \]

### Step 6: Calculate \(a^2 + b^2 + c^2 - ab - ac - bc\)
\[ a^2 + b^2 + c^2 - ab - ac - bc = 3.5558 - (0.6321 + 0.473 + 1.617) = 3.5558 - 2.7221 = 0.8337 \]

### Step 7: Substitute back into the formula
Now substituting back into our expression:
\[ \frac{(0.43)^3 + (1.47)^3 + (1.1)^3 - 3 \cdot 0.43 \cdot 1.47 \cdot 1.1}{0.8337} \]
Since we know that \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)\), we can simplify:
\[ = \frac{3.0 \cdot 0.8337}{0.8337} = 3.0 \]

### Final Answer
Thus, the evaluated expression is:
\[ \boxed{3} \]