Chapter A.2, Problem 56E

Solution

To factor the sum or difference of two cubes $64z^3+27$ .

  $\left(4z+3\right)\left(16z^2-12z+9\right)$

Given:

The expression, $64z^3+27$

Concept Used:

• Like terms are the terms that contain the same variable with similar exponents on it. Constants are like terms as well.
• In any expression, terms are separated by plus or minus signs.
• Coefficients are the constants in front of the variable terms.
• $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$

Calculation:

In order to find the factor the of two cubes first write 27 and $64z^3$ in exponential term then solving the equation by using the difference of cubes formula as shown below

  $27=3^3$

  $64z^3={\left(4z\right)}^3$

  ${\left(4z\right)}^3+3^3$

  $\begin{array}{l}{a}^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\\ {\left(4z\right)}^3+3^3=\left(4z+3\right)\left(16z^2-4z\cdot 3+3^2\right)\\ =\left(4z+3\right)\left(16z^2-12z+9\right)\end{array}$

The original expression now become

  $\left(4z+3\right)\left(16z^2-12z+9\right)$