Q. 402 - Intermediate Algebra

Question

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible. 

$b^3-216$

Step-by-Step Solution

Answer

The Factored form of given polynomial is,

$(b-6)\left(b^2+6b+36\right)$

Step 1. Given Information

We are given a polynomial, 

$b^3-216$

The formula used for factoring using sum and difference of cube is, 

$$\begin{gathered} (a^3-b^3)=(a-b)(a^2+ab+b^2) \\ (a^3+b^3)=(a+b)(a^2-ab+b^2) \end{gathered}$$

Step 2. Factorizing the polynomial

The given polynomial can be also written as,

$\begin{array}{l}{b}^3-216=(b{)}^3-(6{)}^3\end{array}$

Using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$, we get

$$\begin{gathered} b^3-216=(b-6)\left(b^2+b\times 6+6^2\right) \\ {b}^3-216=(b-6)\left(b^2+6b+36\right) \end{gathered}$$

Step 3. Checking the factorization by multiplying

Multiplying the factors, we get 

$$\begin{gathered} (b-6)\left(b^2+6b+36\right)=b^3-216 \\ {b}^3+6b^2+36b-6b^2-36b-216=b^3-216 \\ {b}^3-216=b^3-216 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.