Question

a minus b whole cube plus B minus C whole cube plus c minus a whole cube

Answer 1

( a - b )^3 + ( b - c )^3 + ( c - a )^3

if a + b + c = 0 , then ( a - b )^3 + ( b - c )^3 + ( c - a )^3

sum = a - b + b - c + c -a = 0

( a - b )^3 + ( b - c )^3 + ( c - a )^3 = 3 ( a-b ) (b - c )(c -a) Ans.

Answer 2

Given,

equation given is

$(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$               (1)

To Find,

the value of

$(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$

Solution,

we will use the identity $(a-b)^{3}= a^{3}-b^{3}-3a^{2}b+3ab^{2}$ to find the required value.

so,

$(a-b)^{3}= a^{3}-b^{3}-3a^{2}b+3ab^{2}$            (2)

$(b-c)^{3}= b^{3}-c^{3}-3b^{2}c+3bc^{2}$             (3)

$(c-a)^{3}= c^{3}-a^{3}-3c^{2}a+3ca^{2}$            (4)

now, using the value of equations (2),(3),(4) in equation (1)

the equation (1) becomes,

$a^{3}-b^{3}-3a^{2}b+3ab^{2}$ +$b^{3}-c^{3}-3b^{2}c+3bc^{2}$+$c^{3}-a^{3}-3c^{2}a+3ca^{2}$

= $-3a^{2}b+3ab^{2}$$-3b^{2}c+3bc^{2}$$-3c^{2}a+3ca^{2}$

= $3a^{2}(c-b)+3b^{2}(a-c)+3c^{2}(b-a)$

Hence the value of $(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$ is $3a^{2}(c-b)+3b^{2}(a-c)+3c^{2}(b-a)$ .