Question

If a, b and c are positive integers, then(a – b – c)3 – a3 + b3 + c3 is always divisible by(1) Only (a + b) (2) Only (b + c)(3) 3(a – b) (4) 3(a + b)(b + c)(a + c)​

Answer 1

Answer:

3(a-b)

Step-by-step explanation:

We will deduce the formula of (a-b-c)³.

Now, (a-b-c)³

=(a-b)³+3(a-b)²(-c)+3(a-b)(-c)²+(-c)³

=(a³-b³-3a²b+3ab²)-3(a²+b²-2ab)c+3ac²-3bc²-c³

[Here, we have used the formula of (a+b)²=a²+b²+2ab, and the formula (a-b)³= a³-b³-3a²b+3ab²]

=a³-b³-c³-3a²b+3ab²-3b²c-3bc²+3c²a-3ca²+6abc

=a³-b³-c³-3ab(a-b)+3(a-b)c²-3c(a²+b²)+6abc

=a³-b³-c³-3ab(a-b)+3(a-b)c²-3c{(a-b)²+2ab}+6abc

=a³-b³-c³-3ab(a-b)+3(a-b)c²-3c(a-b)²-6abc+6abc

=a³-b³-c³-3(a-b){ab-c²+c(a-b)}

=a³-b³-c³-3(a-b)(b+c)(a-c)

Taking (a³-b³-c³) to the right side of the equation.

Hence, (a+b+c)³-a³+b³+c³=3(a-b)(b+c)(a-c)

Therefore, (a+b+c)³-a³+b³+c³ is always divisible by 3(a-b).(Answer)

{Note: The formula of (a+b+c)³ is given by

(a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a)

Now, putting b=-b and c=-c, we get the following formula:

(a-b-c)³=a³-b³-c³+3(a-b)(-b-c)(-c+a)

⇒(a-b-c)³-a³+b³+c³=-3(a-b)(b+c)(a-c)}