# If a+b=1, prove that (a2-b2)2=a3+b3-ab

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### 1 Answer

You need to prove that `(a^2-b^2)^2=a^3+b^3-ab` using the information that `a+b = 1,` hence, you should convert the sum of cubes from the right into a product such that:

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

Since `a+b = 1` , hence `a^3+b^3 = (a^2 - ab + b^2)`

You need to substitute `a^2 - ab + b^2` for `a^3+b^3` such that:

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

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

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

You should now convert the difference of squares from the left into a special product such that:

`a^2-b^2 = (a-b)(a+b) =gt (a^2-b^2)^2 = ((a-b)(a+b))^2`

You need to substitute 1 for a+b such that:

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

**Hence, evaluating both sides yields `(a-b)^2 = (a-b)^2` , thus, the equation `(a^2-b^2)^2=a^3+b^3-ab` holds for a+b = 1.**