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Proving the above is easy as long as you are well versed with some of the existing formulas.
(a+b+c)³-a³-b³-c³= [(a+b)+c]-a³-b³-c³
(a+b)c+3c(a+b)(a+b+c)-a³-b³-c³
a³+b³+c+3ab(a+b)+3c(a+b)(a+b+c)-a³-b³-c³
3ab(a+b)+3c(a+b)+3c(a+b)(a+b+c)3(a+b)[ab+c(a+b+c)]
3(a+b)[an+ac+bc+c]3(a+b)[a(b+c)+c(b+c)]
3(a+b)(b+c)(c+a)
Hence proved
Hope this helps you friend
Thanks ✌️✌️
(a+b+c)³-a³-b³-c³= [(a+b)+c]-a³-b³-c³
(a+b)c+3c(a+b)(a+b+c)-a³-b³-c³
a³+b³+c+3ab(a+b)+3c(a+b)(a+b+c)-a³-b³-c³
3ab(a+b)+3c(a+b)+3c(a+b)(a+b+c)3(a+b)[ab+c(a+b+c)]
3(a+b)[an+ac+bc+c]3(a+b)[a(b+c)+c(b+c)]
3(a+b)(b+c)(c+a)
Hence proved
Hope this helps you friend
Thanks ✌️✌️
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