Question

How do you find all solutions to \[{{x}^{3}}+1=0\]?

Answer 1

Hint: From the question we have been asked to find all solutions of \[{{x}^{3}}+1=0\].First of all, by using trial and error method, we have to find one solution. And then using that, we split the given equation into products of a linear expression and quadratic expression. Then by finding the solutions of the quadratic equation, we get the solutions of the given equation.

Complete step-by-step solution:
From the question, we have been given that \[{{x}^{3}}+1=0\]
First of all, let us find a solution by trial and error method.
We can clearly say that \[-1\] is a solution for the given equation by using trial and error methods.
Therefore, \[x=-1\] is one solution for the given equation.
Now, as of process, we have to rewrite the equation.
By rewriting the equation, we get
\[{{x}^{3}}+1=0\]
\[\Rightarrow \left( x+1 \right)\left( {{x}^{2}}-x+1 \right)=0\]
Now, by using zero rule, we have to equate each of the above expressions to zero.
\[x+1=0\]
\[{{x}^{2}}-x+1=0\]
The second equation is in the form of the quadratic equation \[a{{x}^{2}}+bx+c=0\].
By comparing the coefficients of both the equations, we get
\[\begin{align}
  & a=1 \\
 & b=-1 \\
 & c=1 \\
\end{align}\]
Now, we can find the solutions of the above quadratic equation by using the quadratic formula.
Quadratic formula: \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By substituting the values in the formula, we get
\[x=\dfrac{1\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)}\]
\[\Rightarrow x=\dfrac{1\pm \sqrt{-3}}{2}\]
\[\Rightarrow x=\dfrac{1\pm i\sqrt{3}}{2}\]
Therefore, we got all the solutions for the given question.
\[x=-1,x=\dfrac{1\pm i\sqrt{3}}{2}\] are the solutions for the given question.

Note: We should be well aware of the problems like finding solutions for the given polynomial equations. Also, we should be very careful while doing the calculation using the quadratic formula. Also, we should be well aware of the finding roots of quadratic equations. Also, we should be very careful while finding the first solution by using trial and error methods. Similarly we can solve the equation \[{{x}^{4}}-1=0\Rightarrow \left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-1 \right)=0\Rightarrow x=\pm 1,\pm i\].