# How can I demonstrate this is true? 1. `a^3+b^3+c^3-3abc` `=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)` 2. `a^3+b^3+c^3` =`(a+b+c)^3-3(a+b)(b+c)(c+a)` Thank you!

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In terms of eNotes rules, please note that you cannot post multiple questions. These are completely separate questions but can be answered using the same principle as explained above.

**Sources:**

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

Expand the RHS (right-hand side):

RHS= `a^3 +ab^2+ac^2-a^2b-abc-a^2c >`

`` `+a^2b+b^3+bc^2-ab^2-b^2c-abc >`

`+a^2c+b^2c +c^3-abc-bc^2-ac^2gt`

Rearrange and reduce. Note how the positive terms cancel out the negative terms leaving:

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

`therefore = a^3+b^3+c^3-3abc`

**Ans: LHS = RHS. By expanding the right-hand side, you can show that it is equal to the left hand side. **

**Sources:**

Thank you very much!I understand and i'm sorry for this.I didn't know the rules.