Chapter 8.4, Problem 33E

To determine

To expand: and simplify the given expression using binomial theorem.

Answer to Problem 33E

The simplified form is $64y^3-144y^2+108y-27$ .

Explanation of Solution

Given information:

Given expression

  ${\left(4y-3\right)}^3$

Formula used:

From binomial theorem expansion,

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\mathrm{...}{+}_n{C}_r{x}^{n-r}{y}^r+\mathrm{...}+nx{y}^{n-1}+y^n$

And the coefficient of

  $x^{n-r}{y}^r$ is ${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

Calculation:

In the expansion of ${\left(x+y\right)}^n$

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\mathrm{...}{+}_n{C}_r{x}^{n-r}{y}^r+\mathrm{...}+nx{y}^{n-1}+y^n$

Substitute $4y$ for $x$ , $-3$ for $y$ , and 3 for $n$ , we get ${\left(4y-3\right)}^3$

From the above expansion

  $\begin{array}{c}{\left(4y-3\right)}^3{=}_3{C}_0{\left(4y\right)}^3{\left(-3\right)}^0{+}_3{C}_1{\left(4y\right)}^2{\left(-3\right)}^1{+}_3{C}_2{\left(4y\right)}^1{\left(-3\right)}^2{+}_3{C}_3{\left(4y\right)}^0{\left(-3\right)}^3\\ =1{\left(4y\right)}^3{\left(-3\right)}^0+3{\left(4y\right)}^2{\left(-3\right)}^1+3{\left(4y\right)}^1{\left(-3\right)}^2+1{\left(4y\right)}^0{\left(-3\right)}^3\\ =64y^3-144y^2+108y-27\end{array}$