Question

How do you expand ${(c - d)^7}$?

Answer 1

Here we have to expand the expression which is in the power of $7$. We will use the binomial theorem to expand the expression ${(c - d)^7}$. Binomial theorem states that ${(x + y)^7} = \sum\limits_{k = 0}^n {^n{C_k}} {x^{n - k}}{y^k} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + \ldots { + ^n}{C_n}{x^{n - n}}{y^n}$.For calculating $^n{C_k}$ we will use the formula i.e., $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.

Complete step by step answer:
Here we have to expand the expression which is in the power of $7$. As the power of expression is $7$ the expansion of terms becomes very lengthy and tedious to calculate. It can be easily calculated with the help of the Binomial theorem. Binomial theorem is the process of expanding an expression that has been raised to finite power.

Binomial theorem states that we can expand the power ${(x + y)^n}$ into a sum involving terms in the form of $a{x^b}{y^c}$, where exponents $b$ and $c$ are non-negative integers with $b + c = n$ and $a$ is a coefficient of each term which is a specific positive integer depending on $n$ and $b$.
The theorem is given by the formula ${(x + y)^7} = \sum\limits_{k = 0}^n {^n{C_k}} {x^{n - k}}{y^k} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + \ldots { + ^n}{C_n}{x^{n - n}}{y^n}$
We have to expand ${(c - d)^7}$

From binomial theorem we have
${(x + y)^7} = \sum\limits_{k = 0}^n {^n{C_k}} {x^{n - k}}{y^k} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + \ldots { + ^n}{C_n}{x^{n - n}}{y^n}$
Putting $x = c,\,\,y = - d$ and apply a binomial theorem to get the expansion of ${(c - d)^7}$. We get,
$\Rightarrow {(c + ( - d))^7} = \sum\limits_{k = 0}^7 {^7{C_k}} {(c)^{7 - k}}{( - d)^k}$
We can write the above equation as
$\Rightarrow {(c + ( - d))^7} = {c^7}{ + ^7}{C_1}{(c)^{7 - 1}}( - d){ + ^7}{C_2}{(c)^{7 - 2}}{( - d)^2}{ + ^7}{C_3}{(c)^{7 - 3}}{( - d)^3}{ + ^7}{C_4}{(c)^{7 - 4}}{( - d)^4}{ + ^7}{C_5}{(c)^{7 - 5}}{( - d)^5}$ ${ + ^7}{C_6}{(c)^{7 - 6}}{( - d)^6}{ + ^7}{C_7}{(c)^{7 - 7}}{( - d)^7}$

Simplifying the power of the expression. We get,
$\Rightarrow {(c + ( - d))^7} = {c^7}{ + ^7}{C_1}{(c)^6}( - d){ + ^7}{C_2}{(c)^5}{( - d)^2}{ + ^7}{C_3}{(c)^4}{( - d)^3}{ + ^7}{C_4}{(c)^3}{( - d)^4}{ + ^7}{C_5}{(c)^2}{( - d)^5}$ ${ + ^7}{C_6}{(c)^1}{( - d)^6}{ + ^7}{C_7}{( - d)^7}$
Now first we will find the value of terms $^7{C_1},\,{\,^7}{C_2},\,{\,^7}{C_3},\,{\,^7}{C_4},\,{\,^7}{C_5},\,{\,^7}{C_6},\,{\,^7}{C_7}\,\,$ with the help of the formula $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
${ \Rightarrow ^7}{C_1} = \dfrac{{7!}}{{1!(7 - 1)!}} = \dfrac{{7!}}{{1!6!}}$
Expanding $7!$ and cancelling out common terms from both numerator and denominator. We get,
$\Rightarrow \dfrac{{7 \times 6!}}{{1 \times 6!}} = 7$

Now, calculate the value for $^7{C_2}$
${ \Rightarrow ^7}{C_2} = \dfrac{{7!}}{{2!(7 - 2)!}} = \dfrac{{7!}}{{2!5!}}$
Expanding $7!$ and $2!$ and cancelling out common terms from both numerator and denominator.
We get,
$\Rightarrow \dfrac{{7 \times 6 \times 5!}}{{2 \times 1 \times 5!}} = 21$
Now, calculate the value for $^7{C_3}$
${ \Rightarrow ^7}{C_3} = \dfrac{{7!}}{{3!(7 - 3)!}} = \dfrac{{7!}}{{3!4!}}$
Expanding $7!$ and $3!$ and cancelling out common terms from both numerator and denominator. We get,
$\Rightarrow \dfrac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4!}} = 35$

Now, calculate the value for $^7{C_4}$
${ \Rightarrow ^7}{C_4} = \dfrac{{7!}}{{4!(7 - 4)!}} = \dfrac{{7!}}{{4!3!}}$
Expanding $7!$ and $3!$ and cancelling out common terms from both numerator and denominator. We get,
$\Rightarrow \dfrac{{7 \times 6 \times 5 \times 4!}}{{4!\, \times 3 \times 2 \times 1}} = 35$
Now, calculate the value for $^7{C_5}$
${ \Rightarrow ^7}{C_5} = \dfrac{{7!}}{{5!(7 - 5)!}} = \dfrac{{7!}}{{5!2!}}$
Expanding $7!$ and $2!$ and cancelling out common terms from both numerator and denominator.
We get,
$\Rightarrow \dfrac{{7 \times 6 \times 5!}}{{5!\,\, \times 2 \times 1}} = 21$
Now, calculate the value for $^7{C_6}$
${ \Rightarrow ^7}{C_6} = \dfrac{{7!}}{{6!(7 - 6)!}} = \dfrac{{7!}}{{6!1!}}$

Expanding $7!$ and cancelling out common terms from both numerator and denominator.We get,
$\Rightarrow \dfrac{{7 \times 6!}}{{6!\,\, \times 1}} = 7$
Now, calculate the value for $^7{C_7}$
${ \Rightarrow ^7}{C_7} = \dfrac{{7!}}{{7!}} = 1$
Putting these values in the expression. We get,
$\Rightarrow {(c + ( - d))^7} = {c^7} + 7{(c)^6}( - d) + 21{(c)^5}{( - d)^2} + 35{(c)^4}{( - d)^3} + 35{(c)^3}{( - d)^4} + 21{(c)^2}{( - d)^5}$ $+ 7{(c)^1}{( - d)^6} + 1{( - d)^7}$
Simplifying the above equation. We get,
$\therefore {(c - d)^7} = {c^7} - 7{c^6}d + 21{c^5}{d^2} - 35{c^4}{d^3} + 35{c^3}{d^4} - 21{c^2}{d^5} + 7c{d^6} - {d^7}$

Hence, the expansion of ${(c - d)^7} = {c^7} - 7{c^6}d + 21{c^5}{d^2} - 35{c^4}{d^3} + 35{c^3}{d^4} - 21{c^2}{d^5} + 7c{d^6} - {d^7}$.

Note: Some points that should be remember while solving the binomial expansion are that the total number of terms in the expansion ${(x + y)^n}$ is $(n + 1)$ and the sum of exponents is always equal to $n$. $^n{C_1},\,{\,^n}{C_2},\,{\,^n}{C_3},\,{\,^n}{C_4},\, \ldots {,^n}{C_n}\,\,$ are called binomial coefficients and we can also find them with the help of Pascal’s Triangle. The triangle is formed with the help of a simple rule of adding the two numbers above to get the numbers below it.