Chapter 9.5, Problem 84E

To determine

Find the error in given expression.

Answer to Problem 84E

  ${(x-3)}^3$ as ${\left(x+\left(–3\right)\right)}^3$

Explanation of Solution

Given:

  $\begin{array}{cc}& {(x-3)}^3{=}_3{\text{C}}_0{x}^3{+}_3{\text{C}}_1{x}^2(3){+}_3{\text{C}}_{2}x{(3)}^2{+}_3{\text{C}}_3{(3)}^3\dots \dots \dots \dots \dots \mathrm{..}\text{eq}(1)\\ =& 1x^3+3x^{2}3+3\times 3^2+1{(3)}^3\\ =& x^3+9x^2+27x+27\end{array}$

Binomial theorem state that

  ${(a+b)}^n{=}_n{\text{C}}_0{a}^n{+}_n{\text{C}}_1{a}^{n-1}b{+}_n{\text{C}}_2{x}^{n-2}{b}^2{+}_n{\text{C}}_3{a}^{n-3}{b}^3+\dots \dots \dots \dots \dots \mathrm{..}+{\text{C}}_{n-1}a{b}^{n-1}{+}_n{\text{C}}_n{b}^n$ .

Now compare eq(1) to binomial theorem

  $\begin{array}{l}a=x\hfill \\ b=-3\hfill \\ n=3\hfill \end{array}$

Here $b=–3$ , because binomial theorem say expansion rule for ${(a+b)}^n$ not for ${(a-b)}^n$

So let's change negative sign into positive sign using integer law but value of the expression is same that's not changed. So ${(x-3)}^3$ as ${\left(x+\left(–3\right)\right)}^3$

Due to above reason in eq(1) use $\left(–3\right)$ at the place of $3$ .

Sogiven expression look like following expression

  $\begin{array}{l}{(x-3)}^3{=}_3{\text{C}}_0{x}^3{+}_3{\text{C}}_1{x}^2(-3){+}_3{\text{C}}_{2}x{(-3)}^2{+}_3{\text{C}}_3{(-3)}^3\hfill \\ =1x^3+3x^2(-3)+3x{(-3)}^2+1{(-3)}^3\hfill \\ =x^3-9x^2+27x-27\hfill \end{array}$