+6 Hello
We can start by using the identity (x + y + z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x + 3y^2z + 3z^2x + 3z^2y + 6xyz
to expand (ab + ac + bc)^3:
(ab + ac + bc)^3 = a^3b^3 + a^3c^3 + b^3c^3 + 3a^2b^2c(a + b + c) + 3a^2bc^2(a + b + c) + 3ab^2c^2(a + b + c) + 6a^2b^2c^2
Notice that we can simplify the terms involving a + b + c by factoring out the common factor:
(ab + ac + bc)^3 = a^3b^3 + a^3c^3 + b^3c^3 + 3abc(a + b + c)(a^2 + b^2 + c^2) + 6a^2b^2c^2
Substituting a = 1, b = 1, and c = -1,
we get:
(ab + ac + bc)^3 - a^3b^3 = 1 + 1 + 1 + 3(1)(1 - 1)(1^2 + 1^2 + (-1)^2) + 6(1)(1)(-1) - 1 = 3
Therefore,
(ab + ac + bc)^3 - a^3b^3 = 3.
hope so you got your answer please give your feedback.
