Question

Factorise : a3 (b-c)3 + b3 (c-a)3 -c3 (b-a)3

Answer 1

Answer:

a³(b-c)³+b³(c-a)³-c³(b-a)³

=-3a³b³c³(b-c)³(c-a)³(b-a)³

Step-by-step explanation:

Given a³(b-c)³+b³(c-a)³-c³(b-a)³

=[a(b-c)]³+[b(c-a)]³+[-c(b-a)]³

/* We know that,

If x+y+z=0 then

x³+y³+z³=3xyz */

Here ,

x = a(b-c),

y = b(c-a),

z = -c(b-a).

Now,

x+y+z

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

= ab-ac+bc-ab-bc+ac

= 0

=> [a(b-c)]³+[b(c-a)]³+[-c(b-a)]³

= x³+y³+z³

= 3xyz

=3[a(b-c)]³[b(c-a)]³[-c(b-a)]³

=-3a³b³c³(b-c)³(c-a)³(b-a)³

•••♪

Answer 2

Answer:

3abc(b_c)(c_a)(a_b)