Question 3406: can you please help me to factorize the following
1. a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc
2. ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)
3. Use factor theorem to prove ;(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)=24xyz
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! 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
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