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Thread: Factor a^3 + b^3 + c^3 - 3abc

  1. #1
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    Factor a^3 + b^3 + c^3 - 3abc

    .


    Factor the following expression completely so it has integer coefficients:

     a^3 + b^3 + c^3 - 3abc


    And please show your supporting steps.
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  2. #2
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    Re: Factor a^3 + b^3 + c^3 - 3abc

    $\begin{align*}a^3+b^3+c^3-3abc & = a^3-abc+b^3-abc+c^3-abc \\ & = a(a^2-bc)+b(b^2-ac)+c(c^2-ab) \\ & = (a+b+c-b-c)(a^2-bc) + (a+b+c-a-c)(b^2-ac)+(a+b+c-a-b)(c^2-ab) \\ & = (a+b+c)(a^2-bc) -a^2b+b^2c -a^2c+bc^2 + (a+b+c)(b^2-ac)-ab^2+a^2c-b^2c+ac^2+(a+b+c)(c^2-ab)-ac^2+a^2b-bc^2-ab^2 \\ & = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)+\cancel{a^2b-a^2b} + \cancel{ab^2-ab^2} + \cancel{a^2c-a^2c}+\cancel{ac^2-ac^2}+\cancel{b^2c-b^2c}+\cancel{bc^2-bc^2} \\ & = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)\end{align*}$
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  3. #3
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    Re: Factor a^3 + b^3 + c^3 - 3abc

    You could also use Newton's Identities to factor this.
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