Understanding the Algebraic Problem

The question provides us with the sum of three variables, the sum of their squares, and the sum of their cubes. We are asked to find the value of a specific expression involving the product of these three variables, specifically \( \frac{abc}{5} \).

We are given the following equations:

• \( a + b + c = 5 \)
• \( a^2 + b^2 + c^2 = 27 \)
• \( a^3 + b^3 + c^3 = 125 \)

We need to calculate the value of \( \frac{abc}{5} \).

Key Algebraic Identities for Solving

To solve this problem, we will use common algebraic identities that relate the sums of variables, sums of their powers, and their products. The crucial identities are:

1. The square of the sum: \( (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) \)
2. The sum of cubes identity: \( a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab+bc+ca)) \)

These identities will allow us to find the value of \( ab+bc+ca \) first, and then the value of \( abc \).

Step-by-Step Calculation of abc and abc/5

Step 1: Calculate the value of \( ab+bc+ca \)

We use the identity \( (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) \). We substitute the given values:

\( (5)^2 = 27 + 2(ab+bc+ca) \)

\( 25 = 27 + 2(ab+bc+ca) \)

Subtract 27 from both sides:

\( 25 - 27 = 2(ab+bc+ca) \)

\( -2 = 2(ab+bc+ca) \)

Divide by 2:

\( ab+bc+ca = \frac{-2}{2} \)

\( ab+bc+ca = -1 \)

So, the sum of pairwise products \( ab+bc+ca \) is -1.

Step 2: Calculate the value of \( abc \)

Now, we use the identity \( a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab+bc+ca)) \). We substitute the given values and the value of \( ab+bc+ca \) we just found:

\( 125 - 3abc = (5)(27 - (-1)) \)

\( 125 - 3abc = 5(27 + 1) \)

\( 125 - 3abc = 5(28) \)

\( 125 - 3abc = 140 \)

Subtract 125 from both sides:

\( -3abc = 140 - 125 \)

\( -3abc = 15 \)

Divide by -3:

\( abc = \frac{15}{-3} \)

\( abc = -5 \)

So, the product of the variables \( abc \) is -5.

Step 3: Calculate the value of \( \frac{abc}{5} \)

Finally, we need to find the value of \( \frac{abc}{5} \). We substitute the value of \( abc \) we found:

\( \frac{abc}{5} = \frac{-5}{5} \)

\( \frac{abc}{5} = -1 \)

Conclusion

The value of \( \frac{abc}{5} \) is -1.

Let's review the steps and the results obtained:

Revision Table: Algebraic Identities and Calculations

Additional Information: Symmetric Polynomials

The expressions given (\( a+b+c \), \( a^2+b^2+c^2 \), \( a^3+b^3+c^3 \)) and the expressions we calculated (\( ab+bc+ca \), \( abc \)) are examples of symmetric polynomials. A polynomial in variables \( a, b, c \) is symmetric if it remains unchanged when any two variables are swapped.

There are fundamental symmetric polynomials in \( a, b, c \):

• Elementary Symmetric Polynomials:
  1. \( e_1 = a+b+c \)
  2. \( e_2 = ab+bc+ca \)
  3. \( e_3 = abc \)
• Power Sums:
  1. \( p_1 = a \)
  2. \( p_2 = a^2 + b^2 + c^2 \)
  3. \( p_3 = a^3 + b^3 + c^3 \)
  4. And so on (\( p_k = a^k + b^k + c^k \))

Newton's sums are a set of identities that relate the power sums (\( p_k \)) and the elementary symmetric polynomials (\( e_k \)). The identities we used in the solution are derived from or related to Newton's sums for three variables.

• \( p_1 = e_1 \)
• \( p_2 = e_1 p_1 - 2 e_2 \implies e_2 = \frac{e_1 p_1 - p_2}{2} \). This is equivalent to \( ab+bc+ca = \frac{(a+b+c)(a+b+c) - (a^2+b^2+c^2)}{2} = \frac{(a+b+c)^2 - (a^2+b^2+c^2)}{2} \).
• \( p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 \implies 3 e_3 = p_3 - e_1 p_2 + e_2 p_1 \). This is equivalent to \( 3abc = (a^3+b^3+c^3) - (a+b+c)(a^2+b^2+c^2) + (ab+bc+ca)(a+b+c) \). The identity \( a^3+b^3+c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab+bc+ca)) \) is a more direct way to relate \( p_3, e_1, e_2, e_3 \).

Understanding these relationships between power sums and elementary symmetric polynomials is fundamental in the study of polynomial theory and symmetric functions.