Question

question
factor $b^3 - 64z^3$ completely.
answer attempt 1 out of 2

Explanation:

Step1: Identify the formula for difference of cubes

The expression \(b^{3}-64z^{3}\) is a difference of cubes. The formula for factoring the difference of cubes is \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\). Here, we need to identify \(a\) and \(b\) for our expression. We know that \(64z^{3}=(4z)^{3}\), so \(a = b\) and \(b = 4z\) (using the formula notation, note that the variable name in the formula and our expression's variable might overlap, but we'll be careful). So, \(b^{3}-(4z)^{3}\) fits the difference of cubes formula.

Step2: Apply the difference of cubes formula

Using the formula \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\) with \(a = b\) and \(b = 4z\), we substitute these values into the formula.
First, \((a - b)=(b-4z)\).
Then, \(a^{2}=b^{2}\), \(ab=(b)\times(4z) = 4bz\), and \(b^{2}=(4z)^{2}=16z^{2}\). So, \(a^{2}+ab + b^{2}=b^{2}+4bz + 16z^{2}\).
Putting it all together, \(b^{3}-64z^{3}=(b - 4z)(b^{2}+4bz + 16z^{2})\). We check if \(b^{2}+4bz + 16z^{2}\) can be factored further. The discriminant of the quadratic \(Ax^{2}+Bx + C\) (here \(x = b\), \(A = 1\), \(B = 4z\), \(C = 16z^{2}\)) is \(B^{2}-4AC=(4z)^{2}-4\times1\times16z^{2}=16z^{2}-64z^{2}=- 48z^{2}<0\) (for real numbers), so it doesn't factor over the real numbers. So the complete factorization is \((b - 4z)(b^{2}+4bz + 16z^{2})\).

Answer:

\((b - 4z)(b^{2}+4bz + 16z^{2})\)