If `a+b+c=6, a^(2)+b^(2)+c^(2)=14` and `a^(3)+b^(3)+c^(3)=36` then value of abc is equal to

Options

A. 0
B. 4
C. 1
D. 6

Answer

To find the value of \(abc\) given the equations \(a + b + c = 6\), \(a^2 + b^2 + c^2 = 14\), and \(a^3 + b^3 + c^3 = 36\), we can follow these steps:

### Step 1: Use the identity for the square of a sum
We know that:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]
Substituting the known values:
\[ 6^2 = 14 + 2(ab + bc + ca) \]
This simplifies to:
\[ 36 = 14 + 2(ab + bc + ca) \]

### Step 2: Solve for \(ab + bc + ca\)
Rearranging the equation gives:
\[ 36 - 14 = 2(ab + bc + ca) \]
\[ 22 = 2(ab + bc + ca) \]
Dividing by 2:
\[ ab + bc + ca = 11 \]

### Step 3: Use the identity for the sum of cubes
We also know:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \]
Substituting the known values:
\[ 36 - 3abc = 6(14 - 11) \]
This simplifies to:
\[ 36 - 3abc = 6 \cdot 3 \]
\[ 36 - 3abc = 18 \]

### Step 4: Solve for \(abc\)
Rearranging gives:
\[ 36 - 18 = 3abc \]
\[ 18 = 3abc \]
Dividing by 3:
\[ abc = 6 \]

### Final Answer
Thus, the value of \(abc\) is \(6\).

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