Answer:
Hint: (x+y+z)k≡[x+(y+z)]k.
Now use the binomial formula for (a+b)k.Note that just as you can use Pascal's Triangle for binomials, you can use Pascal's Pyramid for trinomials. Otherwise, you can use the Multinomial Theorem as Jp McCarthy suggested
Pascals' Pyramidhope it doesn't seem so weird), I'm looking for a general expanded form of (x+y+z)k,k∈N.
k=1:x+y+z
k=2:x2+y2+z2+2xy+2xz+2yz
k=3:x3+y3+z3+3xy2+3xz2+3yz2+3x2y+3x2z+3y2z+6xyz
k=4:x4+y4+z4+4xy3+4x3y+4xz3+4x3z+4yz3+4y3z+6x2y2+6y2z2+6x2z2+12x2yz+12xy2z+12xyz2
The elements are obviously determined by combinations of their powers, whose sum is always k. I just cannot find the algorithm for element's constants
Step-by-step explanation:
Sana makatulong
