Question

How do you expand \[{\left( {a - b} \right)^3}\]?

Answer 1

Hint: Here we have to expand the given expression. Also we use the binomial expansion formula to find the required answer. On further simplification we get the required answer.

Formula used: \[{(x + y)^n} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2}{ + ^n}{C_3}{x^{n - 3}}{y^3} + ... + {y^n}\]

Complete step-by-step solution:
We have to find the value of \[{\left( {a - b} \right)^3}\].
We will find the value by binomial expansion formula.
According to the theorem, it is possible to expand the polynomial \[{(x + y)^n}\] into a sum involving terms of the form \[a{x^b}{y^c}\] where the exponents b and c are nonnegative integers with \[b + c = n\], and the coefficient a of each term is a specific positive integer depending on n and b.
We know that,
\[{(x + y)^n} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2}{ + ^n}{C_3}{x^{n - 3}}{y^3} + ... + {y^n}\]
For, \[{\left( {x + y} \right)^3}\] we have,
\[ \Rightarrow {\left( {x + y} \right)^3} = {x^3}{ + ^3}{C_1}{x^{3 - 1}}y{ + ^3}{C_2}{x^{3 - 2}}{y^2} + {y^3}\]
We know that,
\[^3{C_1} = 3{;^3}{C_2} = 3\]
Using these values and simplifying we get,
\[ \Rightarrow {\left( {x + y} \right)^3} = {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\]
Now substitute \[x = a,y = - b\] we get,
\[ \Rightarrow {\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\]
Hence, we get
\[ \Rightarrow {\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\]

Thus, \[{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\] is the expansion of the given term.

Note: Binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n + 1 terms of the form
 \[\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right){a^{n - r}}{b^r}\]
In the sequence of terms, the index r takes on the successive values 0, 1, 2, …, n. The coefficients, called the binomial coefficients, are defined by the formula
 \[\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right) = \dfrac{{n!}}{{(n - r)!r!}}\]
In which n! (Called n factorial) is the product of the first n natural numbers 1, 2, 3, …, n (and where 0! is defined as equal to 1).
The coefficients may also be found in the array often called Pascal’s triangle by finding the $r^{th}$ entry of the $n^{th}$ row (counting begins with a zero in both directions). Each entry in the interior of Pascal’s triangle is the sum of the two entries above it.