Question & Answer
QUESTION

If \[{{\left( n \right)}^{3}}-{{\left( n-1 \right)}^{3}}=n+1\], then which of the following can be the value of n?
(a) 0
(b) 1
(c) -1
(d) cannot be determined

ANSWER Verified Verified
Hint: In this question, from the definition of polynomial we can identify it as a cubic polynomial. Now, by using the laws of polynomials we can convert this cubic polynomial into a sum of simplified terms and then can solve further to get the value of n.
\[{{x}^{3}}-{{a}^{3}}={{\left( x-a \right)}^{3}}+3xa\left( x-a \right)\]

Complete step-by-step answer:
Let us look into some of the definitions and results of polynomials.
POLYNOMIAL: An expression of the form \[{{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+......+{{a}_{n-1}}x+{{a}_{n}},\] where \[{{a}_{0}},{{a}_{1}},{{a}_{2}}.......,{{a}_{n-1}},{{a}_{n}}\] are real numbers and n is a non-negative integer, is called a polynomial in the variable x. Polynomials in variable x are generally denoted by \[f\left( x \right)\].
Multiplication of Polynomials: Two polynomials can be multiplied by applying distributive law and simplifying the like terms.
SPECIAL PRODUCTS:
\[\begin{align}
  & \left( x-a \right)\left( {{x}^{2}}+ax+{{a}^{2}} \right)={{x}^{3}}-{{a}^{3}} \\
 & {{x}^{3}}-{{a}^{3}}={{\left( x-a \right)}^{3}}+3xa\left( x-a \right) \\
\end{align}\]
Now, from the given cubic polynomial in n,
\[\begin{align}
  & \Rightarrow {{\left( n \right)}^{3}}-{{\left( n-1 \right)}^{3}}=n+1 \\
 & \Rightarrow {{\left( n-\left( n-1 \right) \right)}^{3}}+3n\left( n-1 \right)\left( n-\left( n-1 \right) \right)=n+1 \\
\end{align}\]
Now, by applying the special product as above mentioned we can rewrite it as:
\[\begin{align}
  & \left[ \because {{x}^{3}}-{{a}^{3}}={{\left( x-a \right)}^{3}}+3xa\left( x-a \right) \right] \\
 & \Rightarrow {{\left( n-n+1 \right)}^{3}}+3n\left( n-1 \right)\left( n-n+1 \right)=n+1 \\
\end{align}\]
Now, on further simplification and rearranging the terms we get,
\[\begin{align}
  & \Rightarrow 1+3n\left( n-1 \right)\left( 1 \right)=n+1 \\
 & \Rightarrow 3n\left( n-1 \right)+1=n+1 \\
 & \Rightarrow 3n\left( n-1 \right)=n \\
\end{align}\]
Now, let us now expand the terms in the above equation.
\[\Rightarrow 3{{n}^{2}}-3n=n\]
Let us now rearrange the terms in the above equation and get,
\[\Rightarrow 3{{n}^{2}}=4n\]
Here, we have two possibilities of n as it is a quadratic equation.
\[\Rightarrow n=0\text{ or }n=\dfrac{4}{3}\]
According to the options given in the question we only consider:
\[\therefore n=0\]
Hence, the correct option is (a).

Note: Instead of using the above special product to expand the given equation we can also expand it with the other cubic polynomial expansion which gives the quadratic expression directly from the expansion. Both the methods give the same result.
\[\left( x-a \right)\left( {{x}^{2}}+ax+{{a}^{2}} \right)={{x}^{3}}-{{a}^{3}}\]
It is important to note that in the above expansion while expanding the corresponding terms we need to be careful about the n term because neglecting the n changes the equation to be simplified which in turn changes the value of the n completely.
The important point here is there are more possible values of n which satisfy the above equation because of the quadratic expression formed by simplification. But, as all the values are not mentioned in the given option so we choose only one possible value given.