Q. 401 - Intermediate Algebra

Question

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

$a^3-125$

Step-by-Step Solution

Answer

The Factored form of given polynomial is, 

$(a-5)\left(a^2+5a+25\right)$

Step 1. Given Information

We are given a polynomial, 

$a^3-125$

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 written as,

$a^3-125=a^3-5^3$

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

$\begin{array}{l}{a}^3-125=(a-5)\left(a^2+a.5+5^2\right)\\ a^3-125=(a-5)\left(a^2+5a+25\right)\end{array}$

Step 3. Checking the factorization by multiplying

Multiplying the factors, we get 

$$\begin{gathered} (a-5)\left(a^2+5a+25\right)=a^3-125 \\ {a}^3+5a^2+25a-5a^2-25a-125=a^3-125 \\ {a}^3-125=a^3-125 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.