How to factorize this question

let f(a, b, c) = a^4 + b^4 + c^4 - 2(ab)^2 - 2(bc)^2 - 2(ac)^2since f(b + c, b, c)
= (b + c)^4 + b^4 + c^4 - 2((b + c)b)^2 - 2(bc)^2 - 2((b + c)c)^2
= b^4 + 4(b^3)(c) + 6(b^2)(c^2) + 4bc^3 + c^4 + b^4 + c^4 - 2b^4 - 4(b^3)(c) - 2(b^2)(c^2) - 2(b^2)(c^2) - 2(b^2)(c^2) - 4bc^3 - 2c^4
= 0
thus, a - b - c is a factor of f(a, b, c)
without loss of generality, b - c - a and c - a - b are the factor of f(a, b, c)since f(- b - c, b, c)
= (- b - c)^4 + b^4 + c^4 - 2((- b - c)b)^2 - 2(bc)^2 - 2((- b - c)c)^2
= b^4 + 4(b^3)(c) + 6(b^2)(c^2) + 4bc^3 + c^4 + b^4 + c^4 - 2b^4 - 4(b^3)(c) - 2(b^2)(c^2) - 2(b^2)(c^2) - 2(b^2)(c^2) - 4bc^3 - 2c^4
= 0
thus, a + b + c is a factor of f(a, b, c)So, a^4 + b^4 + c^4 - 2(ab)^2 - 2(bc)^2 - 2(ac)^2
= k(a + b + c)(a - b - c)(b - c - a)(c - a - b)
by comparing the coefficients, k = 1therefore, a^4 + b^4 + c^4 - 2(ab)^2 - 2(bc)^2 - 2(ac)^2
= (a + b + c)(a - b - c)(b - c - a)(c - a - b)

a⁴ + b⁴ + c⁴ - 2a²b² - 2b²c² - 2a²c²= a⁴ + b⁴ + c⁴ - 2a²b² - 2b²c² + 2a²c² - 4a²c²= (a⁴ - 2a²b² + b⁴) + 2a²c² - 2b²c² + c⁴- 4a²c²= (a² - b²)² + 2c²(a² - b²) + c⁴- 4a²c²= (a² - b² + c²)² - (2ac)²= (a² - b² + c² + 2ac) (a² - b² + c² - 2ac)= ( (a + c)² - b² ) ( (a - c)² - b² ) = (a + b + c) (a - b + c) (a + b - c) (a - b - c)