prove that (a+b+c)power 3-a power 3-b power 3-c power 3=3(a+b) (b+c)(c+a)
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Step-by-step explanation:
Solution
L. H. S= (a+b+c) ³-a³-b³-c³ =====> (i)
Now, Opening formula of (a+b+c)³
As we know that:
(a+b+c)³ = a³+b³+c³+3(a+b) (b+c) (c+a)
So, putting in (i)
(i) ==> L. H. S= a³+b³+c³+3(a+b) (b+c) (c+a) -a³-b³-c³
Therefore, a³+b³+c³ cancels with -a³-b³-c³
And hence we have in last,
3(a+b) (b+c) (c+a) =R. H. S
That is what we have to proof.
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