Question

How do you expand ${\left( {{x^2} + 3} \right)^6}$?

Answer 1

Hint: In this question we have to expand the given expression by using binomial formula which is given by${\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{\left( b \right)^0} + {}^n{C_1}{a^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{a^{n - 2}}{\left( b \right)^2} + {}^n{C_3}{a^{n - 3}}{\left( b \right)^3} + ....... + {}^n{C_n}{a^0}{\left( b \right)^n}$,
Now substituting the $a$ and $b$ values from the given expression in the question and using the formula${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we will get the required result.

Complete step by step answer:
Binomial expansion is given by the formula, ${\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{\left( b \right)^0} + {}^n{C_1}{a^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{a^{n - 2}}{\left( b \right)^2} + {}^n{C_3}{a^{n - 3}}{\left( b \right)^3} + ....... + {}^n{C_n}{a^0}{\left( b \right)^n}$,
Where${}^n{C_0}$,${}^n{C_1}$,${}^n{C_2}$,${}^n{C_3}$……..and${}^n{C_n}$are the binomial coefficients .
So, now given expression is${\left( {{x^2} + 3} \right)^6}$,
Now substituting $a$ and $b$ in the expansion, here $a = {x^2}$, $b = 3$and $n = 6$, then we get,
$ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {}^6{C_0}{\left( {{x^2}} \right)^6}{\left( 3 \right)^0} + {}^6{C_1}{\left( {{x^2}} \right)^{6 - 1}}{\left( 3 \right)^1} + {}^6{C_2}{\left( {{x^2}} \right)^{6 - 2}}{\left( 3 \right)^2} + {}^6{C_3}{\left( {{x^2}} \right)^{6 - 3}}{\left( 3 \right)^3} + {}^6{C_4}{\left( {{x^2}} \right)^{6 - 4}}{\left( 3 \right)^4} + {}^6{C_5}{\left( {{x^2}} \right)^{6 - 6}}{\left( 3 \right)^5}$ $ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {}^6{C_0}{\left( {{x^2}} \right)^6}{\left( 3 \right)^0} + {}^6{C_1}{\left( {{x^2}} \right)^{6 - 1}}{\left( 3 \right)^1} + {}^6{C_2}{\left( {{x^2}} \right)^{6 - 2}}{\left( 3 \right)^2} + {}^6{C_3}{\left( {{x^2}} \right)^{6 - 3}}{\left( 3 \right)^3} + {}^6{C_4}{\left( {{x^2}} \right)^{6 - 4}}{\left( 3 \right)^4}$$ + {}^6{C_5}{\left( {{x^2}} \right)^{6 - 5}}{\left( 3 \right)^5} + {}^6{C_6}{\left( {{x^2}} \right)^{6 - 6}}{\left( 3 \right)^6}$,
We know that n combination r is given by formula${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, and now simplifying the expansion using the formula,
$ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {x^{12}}{\left( 3 \right)^0} + 6{\left( {{x^2}} \right)^5}{\left( 3 \right)^1} + \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}{\left( {{x^2}} \right)^4}\left( 9 \right) + \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}}{\left( {{x^2}} \right)^3}\left( {27} \right)$$ + \dfrac{{6!}}{{4!\left( {6 - 4} \right)!}}{\left( {{x^2}} \right)^2}\left( {81} \right) + \dfrac{{6!}}{{5!\left( {6 - 5} \right)!}}{\left( {{x^2}} \right)^1}\left( {243} \right) + + \dfrac{{6!}}{{6!\left( {6 - 6} \right)!}}{\left( {{x^2}} \right)^0}\left( {729} \right)$,
Now simplifying the expansion we get,
$ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {x^{12}}{\left( 3 \right)^0} + 6\left( {{x^{10}}} \right){\left( 3 \right)^1} + \dfrac{{6!}}{{2!\left( 4 \right)!}}\left( {{x^8}} \right)\left( 9 \right) + \dfrac{{6!}}{{3!\left( 3 \right)!}}\left( {{x^6}} \right)\left( {27} \right)$$ + \dfrac{{6!}}{{4!\left( 2 \right)!}}\left( {{x^4}} \right)\left( {81} \right) + \dfrac{{6!}}{{5!\left( 1 \right)!}}\left( {{x^2}} \right)\left( {243} \right) + \dfrac{{6!}}{{6!\left( 0 \right)!}}\left( {{x^0}} \right)\left( {729} \right)$,
Now applying factorial formula i.e.,$n! = n \times (n - 1) \times \left( {n - 2} \right) \times ........ \times 3 \times 2 \times 1$,we get,
$ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {x^{12}}{\left( 3 \right)^0} + 6\left( {{x^{10}}} \right){\left( 3 \right)^1} + \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6}}{{\left( {1 \times 2} \right)\left( {1 \times 2 \times 3 \times 4} \right)}}\left( {{x^8}} \right)\left( 9 \right) + \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6}}{{\left( {1 \times 2 \times 3} \right)\left( {1 \times 2 \times 3} \right)}}\left( {{x^6}} \right)\left( {27} \right)$
$ + \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6}}{{\left( {1 \times 2 \times 3 \times 4} \right)\left( {1 \times 2} \right)}}\left( {{x^4}} \right)\left( {81} \right) + \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6}}{{\left( {1 \times 2 \times 3 \times 4 \times 5} \right)\left( 1 \right)}}\left( {{x^2}} \right)\left( {243} \right) + \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6}}{{\left( {1 \times 2 \times 3 \times 4 \times 5 \times 6} \right)\left( 1 \right)}}\left( {729} \right)$,
Now simplifying the expansion we get,
$ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {x^{12}}{\left( 3 \right)^0} + 6\left( {{x^{10}}} \right){\left( 3 \right)^1} + \left( {5 \times 3} \right)\left( {{x^8}} \right)\left( 9 \right) + \left( {4 \times 5} \right)\left( {{x^6}} \right)\left( {27} \right) + \left( {5 \times 3} \right)\left( {{x^4}} \right)\left( {81} \right)$
$ + \left( 6 \right)\left( {{x^2}} \right)\left( {243} \right) + \left( 1 \right)\left( {729} \right)$,
Now simplifying by multiplying we get,
$ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {x^{12}} + 6\left( {{x^{10}}} \right)\left( 3 \right) + \left( {15} \right){x^8}\left( 9 \right) + \left( {20} \right)\left( {{x^6}} \right)\left( {27} \right) + \left( {15} \right)\left( {{x^4}} \right)\left( {81} \right)$$ + \left( 6 \right)\left( {{x^2}} \right)\left( {243} \right) + \left( 1 \right)\left( {729} \right)$,
Now final simplification we get,
$ \Rightarrow {\left( {{x^2} + 3} \right)^6} = {x^{12}} + 18{x^{10}} + 135{x^8} + 540{x^6} + 1215{x^4} + 1458{x^2} + 729$.
So, the binomial expansion is ${\left( {{x^2} + 3} \right)^6} = {x^{12}} + 18{x^{10}} + 135{x^8} + 540{x^6} + 1215{x^4} + 1458{x^2} + 729$.

$\therefore $ The expansion form of the given expression ${\left( {{x^2} + 3} \right)^6}$using binomial expansion will be equal to${\left( {{x^2} + 3} \right)^6} = {x^{12}} + 18{x^{10}} + 135{x^8} + 540{x^6} + 1215{x^4} + 1458{x^2} + 729$.

Note: We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.
The binomial theorem is an algebraic method of expanding a binomial expression. Essentially, it demonstrates what happens when you multiply a binomial by itself. For example, consider the expression${\left( {2x + 5} \right)^9}$. It would take quite a long time to multiply the binomial$2x + 5$out nine times. The binomial theorem provides a short cut, or a formula that yields the expanded form of this expression.