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Question

Factorize a3+b3+c3-3abc.


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Solution

Step 1: Factorize the given expression

a3+b3+c3-3abc

use the identity

(x+y)3=x3+y3+3xy(x+y)(x+y)3-3xy(x+y)=x3+y3

put x=b,y=c

a3+b3+c3-3abc

=a3+(c+b)3-3cb(c+b)-3abc=a3+(b+c)3-3bc(a+b+c)

Step 2: Factorize further

using the identity x3+y3=(x+y)(x2+y2-xy)

x=a,y=b+c

a3+(b+c)3=(a+b+c)a2+(b+c)2-a(b+c)

a3+(b+c)3-3bc(a+b+c)=(a+b+c)a2+(b+c)2-a(b+c)-3bc(a+b+c)=(a+b+c)a2+b2+c2+2bc-ab-ac-3bca3+(b+c)3-3bc(a+b+c)=(a+b+c)a2+b2+c2-bc-ab-ac

Hence the factorized form of the polynomial a3+b3+c3-3abc is (a+b+c)(a2+b2+c2-bc-ab-ac).


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