Question

Factorize :-
(a3 +a2+bc)(b-c) + (b3+b2+ca)(c-a) + (c3+c2+ab)(a−b)​

Answer 1

Answer:a³+b³+c³-3abc

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

Step-by-step explanation:

LHS = a³+b³+c³-3abc

= (a³+b³)+c³-3abc

= (a+b)³-3ab(a+b)+c³-3abc

/* By algebraic identity:

x³+y³+3xy(x+y)=(x+y)³

=> x³+y³ = (x+y)³-3xy(x+y) */

= [(a+b)³+c³]-3ab(a+b)-3abc

=[(a+b+c)³-3(a+b)c(a+b+c)]-3ab(a+b+c)

=(a+b+c)[(a+b+c)²-3(a+b)c-3ab]

=(a+b+c)[a²+b²+c²+2ab+2bc+2ca-3ac-3bc-3ab]

=(a+b+c)(a²+b²+c²-ab-bc-ca)

= RHS

Therefore,

a³+b³+c³-3abc

= (a+b+c)(a²+b²+c²-ab-bc-ca)

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Answer 2

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a³+b³+c³-3abc

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

Step-by-step explanation:

LHS = a³+b³+c³-3abc

= (a³+b³)+c³-3abc

= (a+b)³-3ab(a+b)+c³-3abc

/* By algebraic identity:

x³+y³+3xy(x+y)=(x+y)³

=> x³+y³ = (x+y)³-3xy(x+y) */

= [(a+b)³+c³]-3ab(a+b)-3abc

=[(a+b+c)³-3(a+b)c(a+b+c)]-3ab(a+b+c)

=(a+b+c)[(a+b+c)²-3(a+b)c-3ab]

=(a+b+c)[a²+b²+c²+2ab+2bc+2ca-3ac-3bc-3ab]

=(a+b+c)(a²+b²+c²-ab-bc-ca)

= RHS

Therefore,

a³+b³+c³-3abc

= (a+b+c)(a²+b²+c²-ab-bc-ca)