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how to simplify: (a-b)³ + (b-c)³ + (c-a)³ -3(a-b)(b-c)(c-a)?  

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rcprcp | Student, Grade 9 | eNoter

Posted June 30, 2012 at 3:05 PM via web

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how to simplify:

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

 

3 Answers | Add Yours

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 30, 2012 at 3:36 PM (Answer #1)

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You should remember how to expand the cube of binomial such that:

`(a-b)^3 = a^3 - 3a^2*b + 3ab^2 - b^3`

`(a-b)^3 = a^3- b^3 - 3ab(a-b)`

`(b-c)^3 = b^3 - c^3 - 3bc(b-c) `

`(c-a)^3 = c^3 - a^3 - 3ac(c-a) `

Adding `(a-b)^3, (b-c)^3, (c-a)^3`  yields:

`(a-b)^3 + (b-c)^3 + (c-a)^3 = a^3 - b^3 + b^3 - c^3 +c^3 - a^3 - 3(ab(a-b) + bc(b-c) + ac(c-a))`

Reducing like terms yields:

`(a-b)^3 + (b-c)^3 + (c-a)^3 = - 3(ab(a-b) + bc(b-c) + ac(c-a))`

Subtracting 3(a-b)(b-c)(c-a) yields:

`(a-b)^3 + (b-c)^3 + (c-a)^3 - 3(a-b)(b-c)(c-a) = - 3(ab(a-b) + bc(b-c) + ac(c-a)) - 3(a-b)(b-c)(c-a)`

Factoring out -3 yields:

`(a-b)^3 + (b-c)^3 + (c-a)^3 - 3(a-b)(b-c)(c-a) = - 3(ab(a-b) + bc(b-c) + ac(c-a) + (a-b)(b-c)(c-a))`

Opening the brackets to the right side yields:

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

`(a-b)^3 + (b-c)^3 + (c-a)^3 - 3(a-b)(b-c)(c-a) = -3(a^2b - ab^2 + b^2c - bc^2 + ac^2 - a^2c + abc - a^2b - ac^2 + a^2c - b^2c + ab^2 + bc^2 - abc)`

Reducing like terms yields:

`(a-b)^3 + (b-c)^3 + (c-a)^3 - 3(a-b)(b-c)(c-a) = -3*0`

`(a-b)^3 + (b-c)^3 + (c-a)^3 - 3(a-b)(b-c)(c-a) = 0`

Hence, simplifying the given expression yields: `(a-b)^3 + (b-c)^3 + (c-a)^3 - 3(a-b)(b-c)(c-a) = 0.`

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted June 30, 2012 at 5:42 PM (Answer #2)

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To simplify the above expressions, start by expanding the binomials.

Note that we  can expand the (a-b)^3 , (b-c)^3 , and (c-a)^3 using the special product formulas for a cube of a binomial.

The formula is:  (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

  • Expand: (a-b)^3  = a^3 - 3a^2b + 3ab^2 - b^3   
  • Expand: (b-c)^3  = b^3 - 3b^2c + 3bc^2 - c^3
  • Expand: (c-a)^3  = c^3 -  3c^2a + 3ca^2 - a^3  

                                 = c^3 - 3ac^2  + 3a^2c - a^3

Also, expand -3(a-b)(b-c)(c-a) using distributive property.

  • Multiply (a-b) and (b-c).

        (a-b)(b-c) = ab - ac - b^2 - b(-c)

                       = ab - ac - b^2 + bc

  • Then, multiply (ab - ac - b^2 + bc) and (c-a).

        (c - a)(ab - ac - b^2 + bc) = abc - ac^2 - b^2c + bc^2 -a^2b        

                                                 + a^2c + ab^2 - abc

                                              = -ac^2 - b^2c + bc^2 - a^2b +  

                                                  a^2c + ab^2 

  • Then, multiply -3 and (-ac^2 - b^2c + bc^2 - a^2b + a^2c +  ab^2).

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

        = 3ac^2 + 3b^2c - 3bc^2 + 3a^2b - 3a^2c - 3ab^2

 

Next, combine like terms of the above expanded expressions.       

a^3 - 3a^2b + 3ab^2 - b^3

                                + b^3 - 3b^2c + 3bc^2 - c^3

-a^3                                                           +c^3 -3ac^2 +3a^2c                                                             

      + 3a^2b - 3ab^2          + 3b^2c - 3bc^2         +3ac^2 -3a^2c           

---------------------------------------------------------------------

 0   +  0       + 0       + 0     + 0        + 0        + 0   + 0      + 0   


Answer:  0

 

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chaobas | College Teacher | Valedictorian

Posted July 4, 2012 at 2:37 AM (Answer #3)

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here

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

It is in the form of a3 + b3 + c3 – 3abc

were a =(a-b) :b=(b-c) and c=(c-a)


and we have the formula for 

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 –ab –bc –ca)

and we have some condition as

  1. If a3 + b3 + c3 – 3abc = 0 then a + b + c = 0
  2. If a + b + c = 0 then (a3 + b3 + c3 ) = 3abc
  3. If a3 + b3 + c3 – 3abc = 0 then a2 + b2 + c2 = ab + bc + ca



Now in the given equation 

a =(a-b) :b=(b-c) and c=(c-a)

so 

a+b+c = (a-b)+(b-c)+(c=a) = a-b+b-c+c-a = 0

this is the 1st condition so the value of the whole expression is 0

 

(a-b)³ + (b-c)³ + (c-a)³ -3(a-b)(b-c)(c-a) = 0

 

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