Question

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

Answer 1

Answer:

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)

Answer 2

Answer:

we know,

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

Now it is given that : a3 + b3 + c3 = 3abc

So,

a3 + b3 + c3 - 3abc = 0

(a + b + c )(a2 + b2 + c2 -ab - ac -bc) = 0

this means either

(a2 + b2 + c2 -ab - ac -bc) = 0 or (a + b + c ) = 0

(a2 + b2 + c2 -ab - ac -bc) = 0 cannot be zero because:

2a² + 2b² + 2c² -2ab - 2ac - 2bc = 0

a² + b² -2ab + a² +b² +2c² - 2ac -2bc = 0

(a-b)² + a² + c² -2ac + b² + c² -2bc = 0

(a-b)² + (a-c)² + (b-c)² = 0

(a-b)² , (a-c)² ,(b-c)² > = 0

As a≠b≠c , The given value cannot be zero.

This means (a + b + c ) has to be zero.

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