Question

Prove that: (a+b+c)³ = a³+b³+c³+3(a+b)(b+c)(a+c).​

Answer 1

Answer:

(a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a)

we can derive the formulae as following,

(a+b+c)³

=[(a+b)+c]³

=(a+b)³+c³+3(a+b)c[(a+b)+c]

=a³+b³+3ab(a+b)+c³+3(a+b)c[a+b+c]

=a³+b³+c³+3ab(a+b)+3ca(a+b)+3bc(a+b)+3c²(a+b)

=a³+b³+c³+3(a+b)[ab+bc+ca+c²]

=a³+b³+c³+3(a+b)[b(a+c)+c(a+c)]

=a³+b³+c³+3(a+b)(c+a)(b+c)

=a³+b³+c³+3(a+b)(b+c)(c+a)

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Step-by-step explanation:

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Answer 2

please see to the attachment