If `a+b+c=10` and `a^(2)+b^(2)+c^(2)=80` , find the value of `a^(3)+b^(3)+c^(3)-3abc` .

Options

A. `700`
B. `710`
C. `1280`
D. `950`

Answer

To solve the problem, we will use the identity for the sum of cubes:

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

We are given:
1. \( a + b + c = 10 \)
2. \( a^2 + b^2 + c^2 = 80 \)

### Step 1: Find \( ab + ac + bc \)

We can use the identity for the square of the sum of three numbers:

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

Substituting the known values:

\[ 10^2 = 80 + 2(ab + ac + bc) \]

Calculating \( 10^2 \):

\[ 100 = 80 + 2(ab + ac + bc) \]

### Step 2: Rearranging to find \( ab + ac + bc \)

Now, we can rearrange the equation to isolate \( ab + ac + bc \):

\[ 100 - 80 = 2(ab + ac + bc) \]

\[ 20 = 2(ab + ac + bc) \]

Dividing both sides by 2:

\[ ab + ac + bc = 10 \]

### Step 3: Substitute into the identity

Now we can substitute \( a + b + c \), \( a^2 + b^2 + c^2 \), and \( ab + ac + bc \) into the identity:

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

Substituting the values we found:

\[ = 10 \left( 80 - 10 \right) \]

Calculating inside the parentheses:

\[ = 10 \times 70 \]

### Step 4: Final Calculation

Now, we compute the final result:

\[ = 700 \]

Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( \boxed{700} \).

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