If `a+b+c=11,ab+bc+ca=20`, find the value of (i) `a^(2)+b^(2)+c^(2)` and (ii) `a^(3)+b^(3)+c^(3)-3abc`.

Options

A. 81,61
B. 81, 671
C. 27,61
D. 671,81

Answer

To solve the problem, we need to find the values of \( a^2 + b^2 + c^2 \) and \( a^3 + b^3 + c^3 - 3abc \) given the equations:

1. \( a + b + c = 11 \)
2. \( ab + bc + ca = 20 \)

### Step 1: Finding \( a^2 + b^2 + c^2 \)

We can use the identity:

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

Substituting the known values:

\[ a + b + c = 11 \quad \text{and} \quad ab + bc + ca = 20 \]

Calculating \( (a + b + c)^2 \):

\[ (a + b + c)^2 = 11^2 = 121 \]

Now substituting into the identity:

\[ a^2 + b^2 + c^2 = 121 - 2 \times 20 \]

Calculating \( 2 \times 20 \):

\[ 2 \times 20 = 40 \]

Now substituting back:

\[ a^2 + b^2 + c^2 = 121 - 40 = 81 \]

### Step 2: Finding \( a^3 + b^3 + c^3 - 3abc \)

We can use the identity:

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

Substituting the known values:

\[ a + b + c = 11 \]

We already calculated \( a^2 + b^2 + c^2 = 81 \), so now we need \( (a + b + c)^2 - 3(ab + ac + bc) \):

Calculating \( 3(ab + ac + bc) \):

\[ 3(ab + ac + bc) = 3 \times 20 = 60 \]

Now substituting:

\[ (a + b + c)^2 - 3(ab + ac + bc) = 121 - 60 = 61 \]

Now substituting back into the identity:

\[ a^3 + b^3 + c^3 - 3abc = 11 \times 61 \]

Calculating \( 11 \times 61 \):

\[ 11 \times 61 = 671 \]

### Final Answers

(i) \( a^2 + b^2 + c^2 = 81 \)

(ii) \( a^3 + b^3 + c^3 - 3abc = 671 \)