Prove (a-x)is a factor of (x-a)^3+(x-b)^3+(x-c)^3=0
Im not too sure what to do, would appreciate help
Long divide or use synthetic division : )
(x-a)^3 = -(a-x)^3
So that means we have
(a-x)^3 = (x-b)^3 + (x-c)^3
now, using the sum of cubes formula,
(a-x)^3 = (x-b+x-c)((x-b)^2 - (a-b)(x-c) + (x-c)^2)
since the right side is the product of two factors,
(a-x) must divide one of them.
To prove that (a-x) is a factor of the equation (x-a)^3 + (x-b)^3 + (x-c)^3 = 0, we need to show that when we substitute (a-x) into the equation, it will simplify to zero.
So, let's start by substituting (a-x) into the equation:
((a-x)-a)^3 + ((a-x)-b)^3 + ((a-x)-c)^3 = 0
Simplifying this expression, we get:
(-x)^3 + (-(x+b-a))^3 + (-(x+c-a))^3 = 0
Rewriting this in terms of positive powers of x, we have:
(-1)^3 * x^3 + (-(x+b-a))^3 + (-(x+c-a))^3 = 0
Simplifying further, we get:
-x^3 + (a-b+x)^3 + (a-c+x)^3 = 0
Expanding the cubes, we have:
-x^3 + (a^3 - 3a^2b + 3ab^2 - b^3 + 3a^2x - 6abx + 3bx^2 + 3ax^2 - 3bx^2 + 3x^3) + (a^3 - 3a^2c + 3ac^2 - c^3 + 3a^2x - 6acx + 3cx^2 + 3ax^2 - 3cx^2 + 3x^3) = 0
Simplifying this equation, we get:
-a^3 + 3a^2b - 3ab^2 + b^3 - 3a^2x + 6abx - 3bx^2 - 3ax^2 + 3bx^2 - 3x^3 + a^3 - 3a^2c + 3ac^2 - c^3 + 3a^2x - 6acx + 3cx^2 + 3ax^2 - 3cx^2 + 3x^3 = 0
Combining like terms, we have:
-3ab^2 + 3a^2b - 3bx^2 + 3bx^2 - 6abx + 6abx - 3a^2x - 3ax^2 + 3ax^2 - 3ac^2 + 3a^2c - 3cx^2 + 3cx^2 - 3x^3 + 3x^3 - 3ab^2 + 3a^2b - 3bx^2 + 3bx^2 - 6abx + 6abx - 3a^2x - 3ax^2 + 3ax^2 - 3ac^2 + 3a^2c - 3cx^2 + 3cx^2 - 3x^3 + 3x^3 = 0
Simplifying further, we find that all the terms cancel out, so we are left with:
0 = 0
Since we obtained zero as the result, this shows that (a-x) is indeed a factor of the equation (x-a)^3 + (x-b)^3 + (x-c)^3 = 0.