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

Asked on by rishavroy

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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`


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.

We’ve answered 319,859 questions. We can answer yours, too.

Ask a question