Question

(a+b+c)(ab+bc+ac)-abc

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Answer 1

(a + b + c)(ab + bc + ac) - abc

= a²b + abc +a²c + ab² + b²c + abc + abc + bc² + ac²

= (a²b + 2abc + bc²) + (ac² + a²c) + (ab² + b²c)

= b(a² + 2ac + c²) + ac(c + a) + b²(a + c)

= b(a + c) + ac(a + c) + b²(a + c)

= (a + c)[b(a + c) + ac + b²]

= (a + c)(ab + bc + ac + b²)

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

= (a + c)(b + c)(a + b)

Answer 2

\(\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)

\(=a^2b+abc+a^2c+b^2a+b^2c+abc+abc+c^2b+c^2a-abc\)

\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a^2+b^2+2ab-2ab\right)+2abc\)

\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2-2abc+2abc\)

\(=\left(a+b\right)\left(ab+c^2+ca+cb\right)\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)