Question

Factor completely:64 - y^3

Answer 1

From your equation, you can see that you have a difference of two cubes (aka two cubes being subtracted): 64, which is $4^{3}$, and $y^{3}$.

There is rule for the difference of two cubes:
The difference of two cubes is equal to the difference of the cube roots times a binomial, which is the sum of the squares of the roots plus the product of the roots.

That sounds pretty confusing, but it's much easier to understand when put mathematically. Let's say our two cubes are $a^{3}$ and $b^{3}$. The difference of those two cubes is:
$a^{3} - b^{3} = (a - b)( a^{2} + ab + b^{2})$

In our problem, a = 4 (since $a^{3}$ = 64) and b = y (since $b^{3} = y^{3}$. Plug these values into the rule to find the factor of $64 - y^3$:
$64 - y^3 \\ = (4 - y)( 4^{2} + 4y + y^{2}) \\ = (4 - y)( 16 + 4y + y^{2})$

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Answer: $(4 - y)( 16 + 4y + y^{2})$