Question

What is the coefficient of ${x^3}$ in the binomial expansion of ${\left( {4 - x} \right)^9}$ ?

Answer 1

Hint: The given question requires us to find the term independent of x in the binomial expansion ${\left( {4 - x} \right)^9}$ . Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem. To solve the problem, we find the general term of the binomial expansion and then equate the power of x as zero.

Complete step by step answer:
So, we are given the binomial expression ${\left( {4 - x} \right)^9}$. So, we know that the binomial expansion of \[{\left( {x + y} \right)^n}\] is
$\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}} $
So, the binomial expansion of ${\left( {4 - x} \right)^9}$ is
\[\sum\nolimits_{r = 0}^9 {\left( {^9{C_r}} \right){{\left( 4 \right)}^{9 - r}}{{\left( { - x} \right)}^r}} \]
Now, we know that the general term of the binomial expansion of the expression ${\left( {4 - x} \right)^9}$ is \[\left( {^9{C_r}} \right){\left( 4 \right)^{9 - r}}{\left( { - x} \right)^r}\].
Now, we equate the power of x to three so as to find the ${x^3}$ term.
\[r = 3\]
Hence, we substitute the value of r as three. We get, the ${x^3}$ term is
\[\left( {^9{C_3}} \right){\left( 4 \right)^{9 - 3}}{\left( { - x} \right)^3}\]

Simplifying the value of the term, we get,
\[ \Rightarrow \left( {^9{C_3}} \right){\left( 4 \right)^6}\left( { - {x^3}} \right)\]
Using the combination formula \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}\], we get,
\[ \Rightarrow \left( {\dfrac{{9!}}{{3! \times 6!}}} \right){\left( 4 \right)^6}\left( { - {x^3}} \right)\]
Cancelling the common factors in numerator and denominator, we get,
\[ \Rightarrow \left( {\dfrac{{9 \times 8 \times 7 \times 6!}}{{3! \times 6!}}} \right){\left( 4 \right)^6}\left( { - {x^3}} \right)\]
\[ \Rightarrow \left( {\dfrac{{9 \times 8 \times 7}}{6}} \right){\left( 4 \right)^6}\left( { - {x^3}} \right)\]
Simplifying the expression further, we get,
\[ \Rightarrow 84 \times 16 \times 16 \times 16 \times \left( { - {x^3}} \right) = - 344064{x^3}\]

Hence, the coefficient of \[{x^3}\] is \[ - 344064\].

Note: If the value of r comes out to be negative or fractional, we cannot substitute it into the formula for general term as the combination formula \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}\] cannot have negative values in the base. The value of r must be a positive integer less than or equal to n.