Question

23. Prove that (a+b+c)3 – a? - 63 - c = 3(a + b)(b + c)(c + a).​

Answer 1

Answer:

$\small\tt{ANSWER}$

Proving the above is easy as long as you are well versed with some of the existing formula’s and understand how to break the pattern as we progress to obtain the desired result.

Please find below the solution for it:

 Consider:

(a+b+c)3 - a3-b3-c3 = [(a+b)+c]3 – a3-b3-c3

(a+b)3+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+3ab(a+b)+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+c3+3ab(a+b)+3c(a+b)(a+b+c)-a3-b3-c3

3ab(a+b)+3c(a+b)(a+b+c)

3(a+b)[ab+c(a+b+c)]

3(a+b)[ab+ac+bc+c2]

3(a+b)[a(b+c)+c(b+c)]

3(a+b)(b+c)(c+a)

 Hence Proved.