Question

(a+b+c)(a²+b²+c²-ab-bc-ca)​

Answer 1

Answer:

Step-by-step explanation:

(a-b-c)(a²+b²+c²+ab-bc+ca)

=a³-b³-c³-3abc

Step-by-step explanation:

weknow the algebraic identity:

(x+y+z)(x²+y²+z²-xy-yz-zx)

=x³+y³+z³-3xyz--(1)

Now,

wehave,

(a-b-c)(a²+b²+c²+ab-bc+ca)

=[a+(-b)+(-c)][a²+(-b)²+(-c)²-a(-b)-(-b)(-c)-(-c)a]

=a³+(-b)³+(-c)³-3a(-b)(-c)

=a³-b³-c³-3abc/*From (1)

Therefore,

(a-b-c)(a²+b²+c²+ab-bc+ca)

=a³-b³-c³-3abc