Question

Show that (a - b)+(b-c)+(c -a)3 = 3 (a - b) (b -c) (c-a)​

Answer 1

I think that it should be

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

Step-by-step explanation:

Here,

we take , a - b = A,b-c = B , c - a= C

A+B+C = 0

we know that,

{a}^{3}  +  {b}^{3}  +  {c}^{3}   - 3abc = (a + b + c)(  {a}^{2}  +  {b}^{2}  +  {c}^{2}  - ab - bc - ca)

Here , A+B+C = 0

so,

A^3 +B^3 + C^3 = 3 ABC

now we put the values

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

I am done .

Answer 2

Answer:

I think that it should be

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

Step-by-step explanation:

Here,

we take , a - b = A,b-c = B , c - a= C

A+B+C = 0

we know that,

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

Here , A+B+C = 0

so,

A^3 +B^3 + C^3 = 3 ABC

now we put the values

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

I am done .