Question

evaluate. (ii) a + b3 + c3 - 3abc​

Answer 1

Answer:

a3 + b3+ c3− 3abc = (a + b + c)(a2 + b2+ c2− ab − bc − ca)………………………(1)

By using above equation let us consider

a = a, b= -b and c = -c

Then the equation (10 BECOMES

a3 + (-b3) + (-c3) − 3a(-b)(-c) = (a -b -c)(a2 + (-b2) + (-c2) − a(-b) − (-b)(-c) − (-c)a)

a3 -b3 -c3− 3abc = (a -b -c)(a2 +b2 +c2 + ab – bc + ca)

Hence a3 -b3 -c3 − 3abc = (a -b -c)(a2 +b2 +c2 + ab – bc + ca)