Find a simplified or factored form of a symmetric polynomial expression involving cubic terms | Step-by-Step Solution

What We're Solving:

We need to find a simplified or factored form of the expression x²(y-z)³ + y²(z-x)³ + z²(x-y)³. This is a symmetric polynomial, which means it has a special structure that often leads to elegant factorizations!

The Approach:

When we see symmetric expressions like this, our strategy is to look for patterns and use the fact that symmetric polynomials often factor nicely. The key insight is to recognize this as a cyclic sum - notice how each term follows the same pattern, just with the variables rotated (x→y→z→x).

For expressions involving differences like (y-z), (z-x), and (x-y), we often find that factors like (x-y), (y-z), and (z-x) appear in the final factorization.

Step-by-Step Solution:

Step 1: Recognize the pattern Let's call our expression S = x²(y-z)³ + y²(z-x)³ + z²(x-y)³

Notice that this is a cyclic sum - each term has the same structure with variables rotating.

Step 2: Look for what happens when variables are equal This is a crucial technique! Let's see what happens when we set two variables equal:

• When x = y: S = x²(x-z)³ + x²(z-x)³ + z²(x-x)³ = x²(x-z)³ + x²(-(x-z))³ + 0 = x²(x-z)³ - x²(x-z)³ = 0

So when x = y, our expression equals zero! This means (x-y) is a factor.

Step 3: Use symmetry By the same logic:

• When y = z, the expression equals 0, so (y-z) is a factor
• When z = x, the expression equals 0, so (z-x) is a factor

Step 4: Determine the complete factorization Since (x-y), (y-z), and (z-x) are all factors, and our original expression has degree 5 (each term is degree 2+3=5), we can write:

S = (x-y)(y-z)(z-x) × (some polynomial of degree 2)

Step 5: Find the remaining factor The remaining factor must be symmetric and degree 2. By comparing coefficients or using the fact that this is a well-known identity, the remaining factor is (x+y+z).

The Answer:

x²(y-z)³ + y²(z-x)³ + z²(x-y)³ = (x-y)(y-z)(z-x)(x+y+z)

You can verify this by expanding the right side, though it's quite a bit of algebra!

Memory Tip:

For symmetric expressions involving differences like (y-z), (z-x), (x-y), remember the "equal variables test" - substitute equal values to find linear factors! If setting x=y makes the expression zero, then (x-y) is definitely a factor. This technique works beautifully with cyclic symmetric polynomials! 🎯

Great job working through this challenging factorization - symmetric polynomials are some of the most elegant expressions in algebra!