Question

Prove that, (b-c)³+(c-a)³ +(a-b)³ = 3(b-c)(c-a)(a-b)​

Answer 1

Step-by-step explanation:

consider LHS=(b-c)³+(c-a)³+(a-b)³

we know that

(a+b)³=a³+3a²b+3ab²+b³

using this formula

=b³-c³-3b²c+3bc²+c³-a³-3ac²+3a²c+a³-b³-3a²b+3ab²

=-3b²c+3bc²-3ac²+3a²c-3a²b+3ab²---1

now consider RHS=3(b-c)(c-a)(a-b)

=3bc-3ab-3c²+3ac(a-b)

=3abc-3b²c-3a²b+3ab²-3ac²+3bc²+3a²c-3abc

=-3b²c+3bc²-3ac²+3a²c-3a²b+3ab²---(2)

(1)=(2)

hence proved..

Answer 2

Answer:

see in the above picture

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