Question

In the following exercises, expand each binomial ${\left(q\hspace{0.17em}-\hspace{0.17em}4\right)}^3$.

Short answer

The Binomial expansion is $q^3-12q^2+48q-64$.

Step-by-step solution

Step 1. Given Information.

Given Binomial expression is ${\left(q\hspace{0.17em}-\hspace{0.17em}4\right)}^3$.

Step 2. Binomial expansion formula .

The binomial expansion formula is :-

${\left(a+b\right)}^n=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)a^{\left(n-i\right)}{b}^i$

Step 3. Solve Binomial Expansion

Expanding using Binomial Theorem:-

${\left(q\hspace{0.17em}-\hspace{0.17em}4\right)}^3=\sum _{i=0}^3\left(\genfrac{}{}{0ex}{}{3}{i}\right)p^{\left(3-i\right)}{\left(-4\right)}^i$

${\left(q\hspace{0.17em}-\hspace{0.17em}4\right)}^3=\sum _{i=0}^3\left(\genfrac{}{}{0ex}{}{3}{i}\right)p^{\left(3-i\right)}{\left(-4\right)}^i=q^3-12q^2+48q-64$