Question 3549: factorize
1. a(b^4-c^4)+b(c^4-a^4)+c(a^4-b^4)
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! a(b^4-c^4)+b(c^4-a^4)+c(a^4-b^4)
= ca^4 -a^4b + ab^4 -b^4c + bc^4 - c^4a
= ca^4 -b^4c + ab^4 -a^4b - c^4(a-b)
= c(a^4-b^4) -ab(a^3-b^3) - c^4(a-b)
= c(a-b)(a+b)(a^2+b^2) -ab(a-b)(a^2+ab+b^2) - c^4(a-b)
= (a-b)[c(a+b)(a^2+b^2) -ab(a^2+ab+b^2) -c^4]
Next, consider
c(a+b)(a^2+b^2) -ab(a^2+ab+b^2) -c^4
= c(a+b)(a^2+b^2) -ab(a^2+b^2) -a^2b^2 -c^4
= (a^2+b^2)[c(a+b) -ab] -a^2b^2 -c^4
= (a^2+b^2)(c(a+b)-b(a+b))+(a^2+b^2)b^2 -a^2b^2-c^4
= (a^2+b^2)(a+b)(c-b) + b^4-c^4
= (b^2+c^2)(b-c)(b+c) -(b-c)(a^2+b^2)(a+b)
= (b-c)[(b^2+c^2)(b+c) - (a^2+b^2)(a+b)]
= (b-c)[c^3-a^3 + bc^2 + b^2c - ab^2 - a^2b]
= (b-c)[(c-a)(c^2+ca+a^2)+ b^2(c - a) + b(c-a)(c+a)]
= (b-c)(c-a)[(c^2+ca+a^2)+ b^2+ b(c+a)]
= (b-c)(c-a)[a^2+ b^2+c^2+ab+bc+ca]
Hence, a(b^4-c^4)+b(c^4-a^4)+c(a^4-b^4)
= (a-b)(b-c)(c-a)[a^2+ b^2+c^2+ab+bc+ca]
Another way,let f(a,b,c) = a(b^4-c^4)+b(c^4-a^4)+c(a^4-b^4)
f(a,a,c) = a(a^4-c^4)+a(c^4-a^4) = 0.
So, (a-b) is a divisor of f(a,b,c).
Similarly, f(a,b,b) = f(c,b,c) = 0 implies
(b-c) and (c-a) are factors of f(a,b,c).
Hence, (a-b)(b-c)(c-a) are factors of f(a,b,c)
Since degree of f(a,b,c) is 5 and deg of (a-b)(b-c)(c-a) is 3.
Degree of f(a,b,c)/ (a-b)(b-c)(c-a) should be 2.
Assume f(a,b,c)= k(a-b)(b-c)(c-a)(a^2+b^2+c^2- h(ab+bc+ca))
for some constant k and h [why?]
Set a=0,b=1,c=-1
left=1 -(0-1) = 2
right = k(-1)(2)(-1)(2-h(-1)) = 2k(2+h) = 2
k(2+h)=1
Set a=0,b=1,c=2
left hand side = 16 + 2(-1) = 14
right hand side = k(-1)(-1)(2)(5-h(2)) = 2k(5-2h) = 14
k(5-2h) = 7
5-2h = 7, h = -1, k = 1
So, a(b^4-c^4)+b(c^4-a^4)+c(a^4-b^4)
= (a-b)(b-c)(c-a)(a^2+b^2+c^2+ ab+bc+ca)
(same answer)
Kenny
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