Question

Use a Special Factoring Formula to factor the expression

$a^3-b^6$

Short answer

The solution$a^3-b^6=\left(a-b^2\right)\left[a^2+b^4+a{b}^2\right]$

Step-by-step solution

Step 1. Apply the cube of difference formula.

If A and B are any real numbers or algebraic expressions, then

$\begin{array}{l}{\left(A-B\right)}^3=A^3-B^3-3A^{2}B+3A{B}^2\\ \Rightarrow A^3-B^3={\left(A-B\right)}^3+3AB(A-B)\\ \Rightarrow A^3-B^3=(A-B)\left[{\left(A+B\right)}^2+3AB\right]\\ \Rightarrow A^3-B^3=(A-B)\left[A^2+B^2-2AB+3AB\right]\\ \Rightarrow A^3-B^3=(A-B)\left[A^2+B^2+AB\right]\end{array}$

Step 2. Analyze the given expression.

Given$a^3-b^6$

Here

$\begin{array}{l}{A}^3=a^3,B^3=b^6={\left(b^2\right)}^3\\ \Rightarrow A=a,B=b^2\end{array}$

Step 3. Solve the expression.

Substitute the values in the Factoring formula to get:

$\begin{array}{l}{a}^3-b^6=a^3-{\left(b^2\right)}^3\\ \Rightarrow a^3-b^6=\left(a-b^2\right)\left[a^2+{\left(b^2\right)}^2+a\left(b^2\right)\right]\\ \Rightarrow a^3-b^6=\left(a-b^2\right)\left[a^2+b^4+a{b}^2\right]\end{array}$