Chapter 12.6, Problem 45E

To determine

To find: The factor using Binomial Theorem $8a^3+12a^{2}b+6a{b}^2+b^3$

Answer to Problem 45E

  ${\left(2a+b\right)}^3$

Explanation of Solution

Given information:

  $8a^3+12a^{2}b+6a{b}^2+b^3$

Calculation:

According to the Binomial theorem, expansion of ${\left(x+y\right)}^n$ with non-negative integer $n$ is:

  ${\left(x+y\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)x^n{y}^0+\left(\begin{array}{l}n\\ 1\end{array}\right)x^{n-1}{y}^1+\mathrm{.............}+\left(\begin{array}{l}n\\ n-1\end{array}\right)x^1{y}^{n-1}+\left(\begin{array}{l}n\\ n\end{array}\right)x^0{y}^n$

Where $n\ge 0$ , and $\left(\begin{array}{l}n\\ k\end{array}\right)$ denotes the binomial coefficient for $k=0,1,\mathrm{2........}n-1,n$

The number of terms in expansion of ${\left(x+y\right)}^n$ is $n+1$

Given

  $8a^3+12a^{2}b+6a{b}^2+b^3$

Let the factor of $8a^3+12a^{2}b+6a{b}^2+b^3$ be ${\left(x+y\right)}^n$

Here, the number of terms is $4$ .

  $\begin{array}{l}\Rightarrow n+1=4\\ \Rightarrow n=3\end{array}$

Then the factor becomes ${\left(x+y\right)}^3$ .

We know that, ${\left(x+y\right)}^3=x^3+\mathrm{.....}+y^3$

Comparing the first term,

  $\begin{array}{l}{x}^3=8a^3\\ \Rightarrow x=\sqrt[3]{8a^3}\\ \Rightarrow x=\sqrt[3]{{\left(2a\right)}^3}\\ \Rightarrow x=2a\end{array}$

Comparing the last term,

  $\begin{array}{l}{y}^3=b^3\\ \Rightarrow y=\sqrt[3]{b^3}\\ \Rightarrow y=b\end{array}$

Therefore,

  ${\left(x+y\right)}^3={\left(2a+b\right)}^3$

The factor of $8a^3+12a^{2}b+6a{b}^2+b^3$ is ${\left(2a+b\right)}^3$