# How to expand binomials with odd numbered exponents?

I know how to expand when the expression is like: $$a^2-b\qquad\text{ which expands to }\qquad(a+b^{1/2})(a-b^{1/2})$$ or like $$a^4-b\qquad\text{ which expands to }\qquad(a^2+b^{1/2})(a^2-b^{1/2})$$

But what is the method / formula to expand expressions like:

$$a^3-b$$

I would assume its something like the product of a quadratic and a binomial but I am not sure how to get there.

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You can write $$a^3-b=a^3-(b^{1/3})^3=(a-b^{1/3})(a^2+ab^{1/3}+b^{2/3})$$ using the factorization of a difference of cubes.
In general, for any number $n$ we have the factorization $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})$$ (see here). As you can see, there may be further factorization we can do; this equation tells you that $$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$$ but we also know that $$x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)$$ i.e. we can factor the long bit more.