-[b(a-c)-(-b(c-a))]=

Solution for -[b(a-c)-(-b(c-a))]= equation:

Simplifying
-1[b(a + -1c) + -1(-1b(c + -1a))] = 0
-1[(a * b + -1c * b) + -1(-1b(c + -1a))] = 0
-1[(ab + -1bc) + -1(-1b(c + -1a))] = 0

Reorder the terms:
-1[ab + -1bc + -1(-1b(-1a + c))] = 0
-1[ab + -1bc + -1((-1a * -1b + c * -1b))] = 0
-1[ab + -1bc + -1((1ab + -1bc))] = 0
-1[ab + -1bc + (1ab * -1 + -1bc * -1)] = 0
-1[ab + -1bc + (-1ab + 1bc)] = 0

Reorder the terms:
-1[ab + -1ab + -1bc + 1bc] = 0

Combine like terms: ab + -1ab = 0
-1[0 + -1bc + 1bc] = 0
-1[-1bc + 1bc] = 0

Combine like terms: -1bc + 1bc = 0
-1[0] = 0

Multiply -1 * 0
0 = 0

Anything times zero is zero.
0 = 0

Solving
0 = 0

Couldn't find a variable to solve for.

This equation is an identity, all real numbers are solutions.