# squares or cubesa^6-b^6 my tutor said that it must use squares and cubes differences i don't get it

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Using the law of exponentials yields:

`a^6 = a^(2*3) => a^(2*3) = (a^2)^3`

Reasoning by analogy yields:

`b^(2*3) = (b^2)^3`

Converting the difference of cubes into a product yields:

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

You may convert also the difference of squares `a^2 - b^2` into a product, such that:

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

**Hence, converting the difference `a^6 - b^6` into a product, yields **`(a^2)^3 - (b^2)^3 = (a - b)(a + b)(a^4 + (ab)^2 + b^4).`

a^6 - b^6 = (a^3)^2 - (b^3)^2

We'll put a^3 = t and b^3 = u

t^2 - u^2 is a difference of squares

t^2 - u^2 = (t-u)(t+u)

(a^3)^2 - (b^3)^2 = (a^3 - b^3)(a^3 + b^3) difference and sum of cubes.

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

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

So, to solve the problem, we'll apply difference of squares first, then difference and sum of cubes.