If `a + b + c = 15 and (1)/(a) + (1)/(b) + (1)/( c) = (71)/(abc)` then the value of `a^(3) + b^(3) + c^(3) - 3 abc` is

Options

A. 160
B. 180
C. 200
D. 220

Answer

To solve the problem, we start with the given equations:

1. \( a + b + c = 15 \)
2. \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{71}{abc} \)

From the second equation, we can rewrite it using a common denominator:

\[ \frac{bc + ac + ab}{abc} = \frac{71}{abc} \]

By multiplying both sides by \( abc \), we get:

\[ bc + ac + ab = 71 \]

Now we have two equations:

1. \( a + b + c = 15 \) (Equation 1)
2. \( ab + ac + bc = 71 \) (Equation 2)

Next, we need to find \( a^3 + b^3 + c^3 - 3abc \). 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 values from our equations into this identity:

1. \( a + b + c = 15 \)
2. \( ab + ac + bc = 71 \)

Now we calculate \( (a + b + c)^2 \):

\[ (a + b + c)^2 = 15^2 = 225 \]

Next, we substitute into the identity:

\[ a^3 + b^3 + c^3 - 3abc = 15 \left( 225 - 3 \times 71 \right) \]

Calculating \( 3 \times 71 \):

\[ 3 \times 71 = 213 \]

Now substituting this back into our equation:

\[ a^3 + b^3 + c^3 - 3abc = 15 \left( 225 - 213 \right) \]

Calculating \( 225 - 213 \):

\[ 225 - 213 = 12 \]

Now substituting this back:

\[ a^3 + b^3 + c^3 - 3abc = 15 \times 12 \]

Calculating \( 15 \times 12 \):

\[ 15 \times 12 = 180 \]

Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is:

\[ \boxed{180} \]