Question

Prove that the expression is equivalent to zero: a(b+c–bc)–b(c+a–ac)+c(b–a)

Answer 1

Step-by-step explanation:

$a(b+c-bc)-b(c+a-ac)+c(b-a)$

Distribute over the parenthesis:

$(ab+ac-abc)+(-bc-ab+abc)+(bc-ac)\\ab+ac-abc-bc-ab+abc+bc-ac$

Group like terms:

$(ab-ab)+(ac-ac)+(abc-abc)+(cb-cb)$

Anything subtracted from itself is 0.

Simplify:

$0+0+0+0\\0$