Question

Solve \[{{x}^{3}}+3{{x}^{2}}+3x+1=0\]

Answer 1

Hint: We are given a polynomial expression with a degree 3 and we are asked to solve the given expression for the value of ‘x’. As we can see that the given expression looks similar to the expanded form of the algebraic formula of \[{{\left( a+b \right)}^{3}}\], which is, \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\]. So, we will be using this formula and solving the expression and then reducing it to the simplest form possible and finally obtaining the value of ‘x’. Hence, we will have the value of ‘x’.

Complete step by step answer:
According to the given question, we are given a polynomial expression which is of the degree 3 and we are asked to solve the given expression and find the values of ‘x’.
The expression we have is,
\[{{x}^{3}}+3{{x}^{2}}+3x+1=0\]
If we observe the above expression, we can see that it resembles to the expanded form of the algebraic identity of \[{{\left( a+b \right)}^{3}}\], which is, \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\]. We will use this formula in our solution, we have.
We will re-arrange the above expression and we have,
\[\Rightarrow {{x}^{3}}+1+3{{x}^{2}}+3x=0\]
Since, it resembles the identity of \[{{\left( a+b \right)}^{3}}\], so we can write,
\[\Rightarrow {{\left( x+1 \right)}^{3}}=0\]
We can also write the above expression as,
\[\Rightarrow \left( x+1 \right)\left( x+1 \right)\left( x+1 \right)=0\]
We can see that all the brackets are same and so now, we will equate the brackets to zero and e get the values as follows,
\[x=-1,-1,-1\]
Therefore, the values of \[x=-1,-1,-1\].

Note: The above solution gave us three values for ‘x’, since the degree of the given expression is 3 and so the number of zeroes is 3 as well. To check if the value of ‘x’ is correct or not, we simply substitute it back in the given expression and if it comes out to be zero, then the answer is correct else not. We have,
\[{{x}^{3}}+3{{x}^{2}}+3x+1\]
\[\Rightarrow {{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}+3\left( -1 \right)+1\]
\[\Rightarrow -1+3-3+1=0\]
Therefore, the answer we obtained is correct.