Question

The value of ${{\left( 101 \right)}^{3}}$ is
A. 10303001
B. 10030301
C. 1030301
D. 10300301

Answer 1

Hint: We first find the simplification of the given polynomial ${{\left( 101 \right)}^{3}}$ according to the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$. We need to simplify the cubic polynomial of the sum of two numbers as 100 and 1. We already have the identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We then multiply the term $\left( a+b \right)$ on both sides of the identity. We solve the multiplication to find the simplified form of ${{\left( 101 \right)}^{3}}$ by replacing with $a=100;b=1$.

Complete step-by-step solution:
We need to find the simplified form of ${{\left( 101 \right)}^{3}}$. This is the cube of sum of two numbers. We take $a=100;b=1$ for the identity of ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
We need to multiply the term $\left( a+b \right)$ on both side of the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
On the left side of the equation, we get ${{\left( a+b \right)}^{2}}\left( a+b \right)={{\left( a+b \right)}^{3}}$.
On the right side we have $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right)$. We use multiplication and get
$\begin{align}
  & \Rightarrow \left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right) \\
 & ={{a}^{2}}.a+a.{{b}^{2}}+2ab\times a+{{a}^{2}}.b+{{b}^{2}}.b+2ab.b \\
 & ={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \\
\end{align}$
We also can take another form where ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
Now we replace the values for $a=100;b=1$ in the equation ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$.
${{\left( 100+1 \right)}^{3}}={{100}^{3}}+3\times {{100}^{2}}\times 1+3\times 100\times {{1}^{2}}+{{1}^{3}}$
Therefore, the simplified form of ${{\left( 101 \right)}^{3}}$ is
${{\left( 100+1 \right)}^{3}}=1000000+30000+300+1=1030301$
Thus, the final result is ${{\left( 100+1 \right)}^{3}}=1030301$. The correct option is C.

Note: We also can use the binomial theorem to find the general form and then put the value of 3.
We have ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+....+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. We need to find the cube of sum of two numbers. So, we put $n=3$.
${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}+{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}+{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$.
In this way we also simplify the term of ${{\left( a-b \right)}^{3}}$.