prove that (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)no

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rcmath | High School Teacher | (Level 1) Associate Educator

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Easier than proving one of them equal to the other directly, I will multiply and simplify both independtly and prove that they are equal to the same value.

Remember that

`(x-y)^3=x^3-3x^2y+3xy^2-y^3`

So:

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

`a^6-3a^4b^2+3a^2b^4-b^6+b^6-3b^4c^2+3b^2c^4-c^6+`

`c^6+3a^4c^2-3a^2c^4-a^6=`

`-3a^4b^2+3a^2b^4+3a^4c^2-3a^2c^4-3b^4c^2+3b^2c^4=`

`3(-a^4b^2+a^2b^4+a^4c^2-a^2c^4-b^4c^2+b^2c^4)`

After using the difference of square formula, The right hand side of your equation will equal

`3(a^2-b^2)(b^2-c^2)(c^2-a^2)=`

`3(a^2b^2-a^2c^2-b^4+b^2c^2)(c^2-a^2)=`

`3(a^2b^2c^2-a^4b^2-a^2c^4+a^4c^2-b^4c^2+a^2b^4+b^2c^4-a^2b^2c^2)=`

`3(-a^4b^2-a^2c^4+a^4c^2-b^4c^2+a^2b^4+b^2c^4)`

If you rearange the terms in the last factor you will see that they are identical to the result we got form multiplying the left hand side, hence they are equal.