Answer:
k = [tex]\frac{-5}{6}[/tex]
Explanation:
If we assume such a constant exists, then we can solve for it by setting up an equation based on
P(x,y,z)=0
whenever
x+y+z=0.
For example, P(2,-1,-1)=0.
Now we put x = 2, y = -1, z=-1
p(x, y, z) = x⁵ + y⁵ + z⁵ + k(x³+y³+z³)(x²+y²+z²)
p(2,-1,-1) = 2⁵ + (-1)⁵+ (-1)⁵ + k(2³ + (-1)³+ (-1)³)(2² + (-1)²+ (-1)²)
0 = 32 - 1 -1 + k(8-1-1)(4+1+1)
0 = 30 + k(6)(6)
0 = 30 + 36k
solve for k
k = [tex]\frac{-30}{36}[/tex]
k = [tex]\frac{-5}{6}[/tex]
