If a= 5.431, b= 2.121 & c= `-7.552` find `a^(3) +b^(3) + c^(3) + 3abc=` ?

Options

A. 6ab
B. 6abc
C. 1
D. 0

Answer

To solve the problem \( a^3 + b^3 + c^3 + 3abc \) given \( a = 5.431 \), \( b = 2.121 \), and \( c = -7.552 \), we can follow these steps:

### Step 1: Calculate \( a + b + c \)
We first need to find the sum of \( a \), \( b \), and \( c \).

\[ a + b + c = 5.431 + 2.121 - 7.552 \]

Calculating this:

\[ = 7.552 - 7.552 = 0 \]

### Step 2: Apply the identity
Since \( a + b + c = 0 \), we can use the algebraic identity:

\[ a^3 + b^3 + c^3 = 3abc \]

### Step 3: Calculate \( abc \)
Now, we need to calculate \( abc \):

\[ abc = 5.431 \times 2.121 \times (-7.552) \]

Calculating \( 5.431 \times 2.121 \):

\[ = 11.507751 \]

Now multiplying by \( -7.552 \):

\[ abc = 11.507751 \times -7.552 \approx -86.851 \]

### Step 4: Calculate \( a^3 + b^3 + c^3 + 3abc \)
Now, substituting back into the expression:

\[ a^3 + b^3 + c^3 + 3abc = 3abc + 3abc = 6abc \]

Calculating \( 6abc \):

\[ 6abc = 6 \times (-86.851) \approx -521.106 \]

### Final Result
Thus, the value of \( a^3 + b^3 + c^3 + 3abc \) is approximately:

\[ \boxed{-521.106} \]