The problem presented is:

Given: $a + b + c = 5,$ $a^2 + b^2 + c^2 = 13,$ $a^3 + b^3 + c^3 = 47.$ Find the value of $abc$.

Solution:

1. Square the sum $a + b + c = 5$: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ Substituting the given values: $5^2 = 13 + 2(ab + bc + ca)$ $25 = 13 + 2(ab + bc + ca)$ $2(ab + bc + ca) = 12 \quad \Rightarrow \quad ab + bc + ca = 6$

2. Use the identity for the sum of cubes: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ Substituting the given values and results obtained: $47 - 3abc = 5 \times (13 - 6)$ $47 - 3abc = 5 \times 7$ $47 - 3abc = 35$ $3abc = 47 - 35 = 12$ $abc = \frac{12}{3} = 4$

Final Answer:

The value of $abc$ is $4$.

Would you like more details or have any questions? Here are some related questions:

1. How do identities for powers of sums work for more variables?
2. Can this method be applied to any system of equations involving sums of powers?
3. What is the significance of symmetric polynomials in solving such equations?
4. How can these methods be extended to larger systems, e.g., four variables?
5. Can we derive similar identities for higher powers?

Tip: When dealing with symmetric equations, consider using identities and expanding expressions to relate different powers of sums and products.