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# Prove (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a-b)(b-c)(c-a)

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Prove (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a-b)(b-c)(c-a) without expanding using the Multivariable Factor Theorem

Aug 21, 2018

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(a - b)^3 + (b - c)^3  + ( c -a)^3  =

[( a - b ) + (b - c)] [ ( a - b)^2 - ( a - b)(b - c) + ( b - c)^2 ] + (c -a)^3 =

( a - c) [ a^2 - 2ab + b^2  - [ ab - b^2 - ac + bc] + b^2 - 2bc + c^2 ] + (c -a)^3 =

(a - c)  [ a^2 - 2ab + b^2  - ab + b^2 + ac- bc  + b^2 - 2bc + c^2 ]  + (c -a)^3  =

(a - c) [ a^2 - 3ab + 3b^2 + ac - 3bc  + c^2 ]  + (c - a)^3 =

(a - c) [ a^2 - 3ab + 3b^2 + ac - 3bc  + c^2 ] + ( c - a)^3]=

(a - c) [ a^2 - 3ab + 3b^2 + ac - 3bc  + c^2 ] + [ (c - a) (c - a)^2 ]  =

- (c - a) [ a^2 - 3ab + 3b^2 + ac - 3bc + c^2 ] [ (c - a)( c^2 - 2ac + a^2 ]  =

(c - a) [ -a^2 + 3ab - 3b^2 - ac + 3bc - c^2 + c^2 - 2ac + a^2 ] =

(c -a) [ 3ab  - 3b^2 - 3ac + 3bc] =

(c - a) [ 3ab - 3ac - 3b^2 + 3bc ] =

(c - a) [ 3a ( b - c) - 3b ( b - c) ]  =

(c - a) [3(a - b)) ( b - c)]  =

3(a - b)(b - c) (c - a)

Aug 21, 2018