Question

Factor using the binomial theorem? \[8{a^3} + 12{a^2}b + 6a{b^2} + {b^3}\].

Answer 1

Hint: We know that the binomial expansion of \[{\left( {x + y} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + {}^n{C_3}{x^{n - 3}}{y^3} + ..... + {}^n{C_n}{x^0}{y^n}\] .In order to solve this question, we will first find the value of \[n\] using the basic concept of binomial expansion that the total number of terms in the expansion is \[\left( {n + 1} \right)\]. After that we will compare the first and the last terms of the given expansion with the first and the last terms of the general form of the binomial expansion respectively to find the value of \[x\] and \[y\] . And hence we will get the required factor of the given expansion.

Complete step by step answer:
We have given the expansion as \[8{a^3} + 12{a^2}b + 6a{b^2} + {b^3}\]
Now we know that the binomial expansion of
\[{\left( {x + y} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + {}^n{C_3}{x^{n - 3}}{y^3} + ..... + {}^n{C_n}{y^n}\]
We also know that the total number of terms in the expansion is \[\left( {n + 1} \right)\]
So, in the given expansion we have total \[4\] number of terms,
Therefore, we get
\[n + 1 = 4\]
On subtracting \[1\] from both the sides, we get
\[ \Rightarrow n = 3\]
So, the given expansion can be written as,
\[{\left( {x + y} \right)^3} = 8{a^3} + 12{a^3}b + 6a{b^2} + {b^3}\]
Now we will compare the first and the last terms of the given expansion with the first and the last terms of the general form of the binomial expansion respectively to find the value of \[x\] and \[y\]
So, our first term of the given expansion is \[8{a^3}\]
And first term of the general form of the binomial expansion is \[{}^n{C_0}{x^n}\]
Therefore, on comparing we get
\[{}^n{C_0}{x^n} = 8{a^3}\]
Putting value of \[n\] we have
\[{}^n{C_0}{x^3} = 8{a^3}\]
Now we know that
\[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \cdot r!}}\]
So, \[{}^n{C_0} = \dfrac{{n!}}{{\left( {n - 0} \right)! \cdot 0!}} = 1\]
Therefore, we get
\[{x^3} = 8{a^3}\]
Now we can write \[8{a^3} = {\left( {2a} \right)^3}\]
Therefore, we get
\[{x^3} = {\left( {2a} \right)^3}\]
\[ \Rightarrow x = 2a\]
Now our last term of the given expansion is \[{b^3}\]
And last term of the general form of the binomial expansion is \[{}^n{C_0}{y^n}\]
Therefore, on comparing we get
\[{}^n{C_n}{y^n} = {b^3}\]
Putting value of \[n\] we have
\[{}^n{C_n}{y^3} = {b^3}\]
On solving, we get
\[{y^3} = {b^3}\]
\[ \Rightarrow y = b\]
Therefore, on substituting the values, we get our given expansion as
\[{\left( {2a + b} \right)^3} = 8{a^3} + 12{a^2}b + 6a{b^2} + {b^3}\]
Hence, \[{\left( {2a + b} \right)^3}\] is the factor of the given expansion.

Note:
We can also check whether we get the correct factor or not.
Note that \[8{a^3} = {\left( {2a} \right)^3}\] and \[{b^3}\] are both perfect cubes, and this polynomial is homogeneous of degree \[3\] .
So, let’s expand \[{\left( {2a + b} \right)^3}\] and see if it is equal.
we know that
\[{\left( {x + y} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + {}^n{C_3}{x^{n - 3}}{y^3} + ..... + {}^n{C_n}{x^0}{y^n}\]
Here, \[x = 2a,{\text{ }}y = b\] and \[n = 3\]
On substituting we get
\[{\left( {2a + b} \right)^3} = {}^3{C_0}{\left( {2a} \right)^3} + {}^3{C_1}{\left( {2a} \right)^{3 - 1}}{b^1} + {}^3{C_2}{\left( {2a} \right)^{3 - 2}}{b^2} + {}^3{C_3}{\left( {2a} \right)^{3 - 3}}{b^3}\]
On solving, we get
\[ \Rightarrow {\left( {2a + b} \right)^3} = {}^3{C_0}8{a^3} + {}^3{C_1}4{a^2}b + {}^3{C_2}2a{b^2} + {}^3{C_3}{b^3}\]
Now we know that
\[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \cdot r!}}\]
Therefore,
\[{}^3{C_0} = \dfrac{{3!}}{{\left( {3 - 0} \right)! \cdot 0!}} = \dfrac{{3!}}{{0! \cdot 3!}} = 1\]
\[{}^3{C_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)! \cdot 1!}} = \dfrac{{3!}}{{2!}} = 3\]
\[{}^3{C_2} = \dfrac{{3!}}{{\left( {3 - 2} \right)! \cdot 2!}} = \dfrac{{3!}}{{1! \cdot 2!}} = 3\]
\[{}^3{C_3} = \dfrac{{3!}}{{\left( {3 - 3} \right)! \cdot 3!}} = \dfrac{{3!}}{{0! \cdot 3!}} = 1\]
On substituting, we get
\[{\left( {2a + b} \right)^3} = 8{a^3} + 3\left( {4{a^2}b} \right) + 3\left( {2a{b^2}} \right) + {b^3}\]
\[ \Rightarrow {\left( {2a + b} \right)^3} = 8{a^3} + 12{a^2}b + 6a{b^2} + {b^3}\]
Thus, we get the same result
Hence, the calculated factor of the given expansion is correct.