Question 3406
1. a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc 
  = a^2 b+ a^2c + b^2 c+ ab^2+c^2a+bc^2+2abc
  = a(b^2+ 2bc+c^2) + a^2 b+ a^2c + b^2 c +bc^2
  = a(b+c)^2 +  a^2(b+c) + bc(b+c)
  = (b+c) [ a^2 +a(b+c)+ bc]
  = (b+c) (a +b)(c+a)
2. ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) 
  = ab(a+b)(a-b)+ b^3c - bc^3 + c^3a - ca^3
  = ab(a+b)(a-b)+ b^3c - ca^3 - bc^3 + c^3a 
  = ab(a+b)(a-b) - c(a^3 - b^3) + c^3(a- b)
 [Use A^3 - B^3 = (A-B)(A^2+AB+B^2) ]
  = (a-b) [ab(a+b) - c(a^2 + ab+b^2) + c^3]
  = (a-b) [(a^2b - c a^2) + (ab^2 - abc) - (b^2c - c^3)]
  = (a-b) [a^2(b - c ) + ab(b - c) - c(b^2 - c^2)]
  = (a-b) (b - c)[a^2 + ab - c(b + c)]
  = (a-b) (b - c)[a^2 -c^2 + b(a- c)]
 = (a-b) (b - c)(c- a)(a+b+c)
  
3. Use factor theorem to prove ;
(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz

 (x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3
 [Use A^3 - B^3 = (A-B)(A^2+AB+B^2) and
   A^3 + B^3 = (A+B)(A^2-AB+B^2) ]
 = [(x+y+z)-(y+z-x)][(x+y+z)^2+(x+y+z)(y+z-x)+(y+z-x)^2]
  -[(z+x-y)+(x+y-z)][(z+x-y)^2-(z+x-y)(x+y-z)+(x+y-z)^2]
 = 2x[(x+y+z)^2+(x+y+z)(y+z-x)+(y+z-x)^2]
  -2x[(z+x-y)^2-(z+x-y)(x+y-z)+(x+y-z)^2]
 = 2x[(x+y+z)^2+(x+y+z)(y+z-x)+(y+z-x)^2 -(z+x-y)^2+(z+x-y)(x+y-z)-(x+y-z)^2] 
 = 2x[(x+y+z)^2 -(z+x-y)^2 
  + (y+z)^2-x^2+ x^2 - (y-z)^2 
  + (y+z-x)^2 -(x+y-z)^2 ]
 = 2x[(x+y+z)^2 -(z+x-y)^2 
  + 4yz 
  + (y+z-x)^2 -(x+y-z)^2 ]
 = 2x[(2x+2z)(2y)  + 4yz  + 2y (-2x+2z) ]
 = 2x * 12yz
 = 24zyz. 

 Another (better) way:
 2) Let f(a,b,c) = ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) 
   Since when a=b, f(a,a,c) = ab(a^2-a^2)+ac(a^2-c^2)+ca(c^2-a^2) 
                    = ca^3-c^3a+c^3a-ca^3 = 0 .
   Similarly, when b=c or c=a, f(a,b.c) = 0.
   
   Hence, (a-b),(b-c),and (c-a) are factors of f(a,b,c).
   By direct division, f(a,b,c)/[(a-b)(b-c) (c-a) = a+b +c.
   So, we obtain f(a,b,c) = (a-b)(b-c)(c-a)(a+b+c)    
   
 3) Let g(x,y,z) = (x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3.
    When x = 0, g(0,y,z) = (y+z)^3-(y+z)^3-(z-y)^3-(y-z)^3
                         = (y-z)^3-(y-z)^3 = 0.
   Similarly, when y=0, g(x,0,z) = 0 amd
   when z=0, g(x,y,0) = 0.
   Hence, x,y and z are factors of g(x,y,z).
   g(z,y,z) and xyz are of the same degree, 3.
   So, g(z,y,z)  = kxyz for some constant k.
   Let x=y=z=1, g(x,y,z) = 27 -1 -1-1 = 24 = k(1)^3 = k,
   so, k = 24 and we obtain g(z,y,z) = 24 xyz.


 Try to read carefully about every step.

 Kenny