Question

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

Answer 1

Here is your answer...
Hope this will help you.....

Answer 2

Answer:

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

Step-by-step explanation:

RHS = (a+b+c)(a²+b²+c²-ab-bc-ca)

=a(a²+b²+c²-ab-bc-ca)+b(a²+b²+c²-ab-bc-ca)+c(a²+b²+c²-ab-bc-ca)

= a³+ab²+ac²-a²b-abc-ca²

+a²b+b³+bc²-ab²-b²c-abc

+a²c+b²c+c³-abc-bc²-c²a

/* after cancellation, we get

= a³+b³+c³-3abc

= LHS

Therefore,

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

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