Question

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

Answer 1

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

Step-by-step explanation:

as we know thay

(x+y+z)³ = x³ + y³ + z³ + 3(x+y)(y+z)(z+x)

putting x = a

y = -b

z = -c

(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)

Taking - out of (-b -c)

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

or taking - inisde a- c

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

hence

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

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

Answer 2

Answer:  The given expression is divisible by 3.

Step-by-step explanation:  Given that a, b and c are positive integers.

We have by expanding the given expression that

$(a-b-c)^3-a^3+b^3+c^3\\\\=(a-b-c)^2(a-b-c)-a^3+b^3+c^3\\\\=(a^2+b^2+c^2-2ab+2bc-2ca)(a-b-c)-a^3+b^3+c^3\\\\=a^3+ab^2+ac^2-a^2b+2abc-2ca^2-2a^2b-b^3-bc^2+2ab^2-2b^2c+2abc-a^2c-b^2c-c^3+2abc-2bc^2+2c^2a-a^3+b^3+c^3\\\\=3ab^2+3ac^2+6abc-3ca^2-3a^2b-3bc^2-3b^2c\\\\=3(ab^2+ac^2+2abc-ca^2-a^2b-bc^2-b^2c),$

which is divisible by 3.

Hence, the given expression is divisible by 3.