Question

Simplify the given expression $3x\left( {{x}^{2}}+1 \right)-\left[ 2x\left( {{x}^{2}}+x-1 \right)+1 \right]-{{x}^{2}}$
\[\begin{align}
  & A.{{x}^{3}}-3{{x}^{2}}+x+1 \\
 & B.{{x}^{3}}-3{{x}^{2}}+5x-1 \\
 & C.{{x}^{3}}+{{x}^{2}}-5x+1 \\
 & D.{{x}^{3}}+3{{x}^{2}}+5x-1 \\
\end{align}\]

Answer 1

Hint: To solve this question, we will assume variables for all given terms separately as:
\[\begin{align}
  & t=3x\left( {{x}^{2}}+1 \right) \\
 & s=2x\left( {{x}^{2}}+x-1 \right)+1 \\
 & \text{and }r={{x}^{2}} \\
\end{align}\]
Simplify then using power of multiplication of x formula given as below:
\[{{x}^{a}}{{x}^{b}}={{x}^{a+b}}\]
Finally, we will calculate the value of t-s-r to get the required result.

Complete step by step answer:
We are given the expression as:
\[3x\left( {{x}^{2}}+1 \right)-\left[ 2x\left( {{x}^{2}}+x-1 \right)+1 \right]-{{x}^{2}}\]
We have to simplify this term. If we multiply a power to x to another power x then, it is of the form.
\[{{x}^{a}}{{x}^{b}}={{x}^{a+b}}\]
Where a and b are powers of x.
Consider \[t=3x\left( {{x}^{2}}+1 \right),s=2x\left( {{x}^{2}}+x-1 \right)+1\text{ and }r={{x}^{2}}\]
So, we have to calculate the value of t - s - r.
Using the above stated power of x multiplication in t, we get:
\[\begin{align}
  & t=3x\left( {{x}^{2}}+1 \right) \\
 & t=3{{x}^{1+2}}+3x \\
 & t=3{{x}^{3}}+3x\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
Now, again using above stated power of x multiplication in s we get:
\[\begin{align}
  & s=2x\left( {{x}^{2}}+x-1 \right)+1 \\
 & s=2{{x}^{1+2}}+2{{x}^{1+1}}-2x+1 \\
 & s=2{{x}^{3}}+2{{x}^{2}}-2x+1\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
Now, finally using the same above stated power of x multiplication in r we get:
\[r={{x}^{2}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}\]
Then finally we have to compute t - s - r.
Substituting the values of t, s and r from equation (i), (ii) and (iii) in above we get:
\[\begin{align}
  & t-s-r=3{{x}^{3}}+3x-\left( 2{{x}^{3}}+2{{x}^{2}}-2x+1 \right)-{{x}^{2}} \\
 & \Rightarrow 3{{x}^{3}}+3x-2{{x}^{3}}-2{{x}^{2}}+2x-1-{{x}^{2}} \\
 & \Rightarrow {{x}^{3}}\left( 3-2 \right)-2{{x}^{2}}-{{x}^{2}}+3x+2x-1 \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}+5x-1 \\
\end{align}\]
Therefore, the value of expression $3x\left( {{x}^{2}}+1 \right)-\left[ 2x\left( {{x}^{2}}+x-1 \right)+1 \right]-{{x}^{2}}$ is ${{x}^{3}}-3{{x}^{2}}+5x-1$

So, the correct answer is “Option B”.

Note: Another method to solve this question can be directly solved before assuming any variable. The solution is as below:
Consider $3x\left( {{x}^{2}}+1 \right)-\left[ 2x\left( {{x}^{2}}+x-1 \right)+1 \right]-{{x}^{2}}$ using ${{x}^{a}}{{x}^{b}}={{x}^{a+b}}$ we get:
\[\begin{align}
  & \Rightarrow 3{{x}^{3}}+3x-\left( 2{{x}^{3}}+2{{x}^{2}}-2x+1 \right)-{{x}^{2}} \\
 & \Rightarrow 3{{x}^{3}}+3x-2{{x}^{3}}-2{{x}^{2}}+2x-1-{{x}^{2}} \\
 & \Rightarrow {{x}^{3}}+5x-2{{x}^{2}}-{{x}^{2}}-1 \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}+5x-1 \\
\end{align}\]
Which is option B. So, the answer is the same.