How can a ^ 3-B ^ 3 + C ^ 3 + A ^ 2 * B-A * B ^ 2 + BC ^ 2-B ^ 2 * c-abc = 0 be reduced to (a + B + C) (a ^ 2 + C ^ 2-B ^ 2-ac) = 0 How to simplify the process
A ^ 3 - B ^ 3 + C ^ 3 + A ^ 2B - AB ^ 2 + BC ^ 2 - B ^ 2C - ABC = (a ^ 3 + A ^ 2B + A ^ 2C) - A ^ 2C - B ^ 3 - AB ^ 2 + BC ^ 2 - B ^ 2C + C ^ 3 = a ^ 2 (a + B + C) - A ^ 2C - B ^ 3 - AB ^ 2 + BC ^ 2 - B ^ 2C - ABC
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