Question

How do you expand ${\left( {x - 1} \right)^3}$?

Answer 1

Hint:
In the given question, we have given two terms, one is variable $x$ and another is $1$. Here $1$ is subtracted from $x$, and has a cubic. This type of question is solved by using the identity. We have the identity \[{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3{a^2}b(a - b)\]. We have to just identify the $a$ and $b$ and put the value of $a$ and $b$ respectively. After solving, we will get the answer.

Complete step by step solution:
Step1: We have given ${\left( {x - 1} \right)^3}$. Compare it with ${\left( {a - b} \right)^3}$
So, we get the place of $a$ has $x$ and place $b$ has $1$.
That is $a = x,{\text{ }}b = 1$
Using the identity ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3{a^2}b + 3ab$

Step2: Therefore, after but the value of $a$ and $b$
we get
\[{\left( {x - 1} \right)^3} = {\left( x \right)^3} - {\left( 1 \right)^3} - 3\left( {{x^2}} \right)1 + 3\left( x \right){\left( 1 \right)^2}\]
If we multiply $x$ three times, we get \[ = {x^2}\] and multiply three times, we get $1$ .
Therefore, we get
\[{\left( {x - 1} \right)^3} = {x^3} - 1 - 3{x^2} + 3x\]
Here, te multiplication of $3$, ${x^2}$ and $1$ is $3{x^2}$ and the multiplication of $3,x{\text{ and }}{(1)^2}{\text{ is }}3x$ .

Note:
A polynomial of one term is called a monomial. A polynomial of two terms is called a binomial. A polynomial of three terms is called a trinomial. A polynomial of degree three is called a cubic polynomial. Every linear polynomial in one variable has a unique zero, a non zero constant polynomial have no zero and every real number is a zero of the zero polynomial $p(x)$n one variable $x$ is an algebraic expression in $x$ of the form \[f(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ............. + {a_1}{x^{ - 1}}{a_0}x\].