how do you solve ${x^3} - x = 1?$
Answer
592.5k+ views
Hint: In such a type of question we have to find the value of x. Here, the power of x is $3$. So there are $3$ values of x we have to find. So for that, we will simplify the given equation. And after that, we will find values of x.
Complete step by step answer:
For solving such types of questions you first have to find the power of a given equation. If the power of any equation is n then that equation has n roots.
Here, the power of this equation is $3$. So these equations have $3$ roots.
Now, we have to find the $3$ roots of these equations for solving this question.
So for that, we will simplify these equations.
For that first, we will divide and multiply these equations on both sides by x.
So after dividing and multiplying this equations both side by x, we get,
$ \Rightarrow x({x^2} - 1) = 1$
Now, suppose if $a \times b = 1$ then, there are two possibilities:
$\begin{gathered}
\Rightarrow a = 1{\text{ and }}b = 1 \\
\Rightarrow a = - 1{\text{ and }}b = - 1 \\
\end{gathered} $
So, here $a = x$and $b = {x^2} - 1$
$\begin{gathered}
\Rightarrow x = 1{\text{ and }}{{\text{x}}^2} - 1 = 1 \\
\Rightarrow x = - 1{\text{ and }}{{\text{x}}^2} - 1 = - 1 \\
\end{gathered} $
Now, we will solve of these question first for$x = 1{\text{ }}$;
$\begin{gathered}
\Rightarrow {{\text{x}}^2} - 1 = 1 \\
\Rightarrow {x^2} = 2 \\
\Rightarrow x = \pm \sqrt 2 \\
\end{gathered} $
So, for first possibility x have $3$ roots which is given below:
$x = 1{\text{ ,}} \pm \sqrt 2 $
Now, for second possibility:
If $x = - 1{\text{ }}$
Then we get,
$
\Rightarrow {{\text{x}}^2} - 1 = - 1 \\
\Rightarrow {x^2} = 0 \\
\Rightarrow {\text{not possible}} \\
$
Because when we substitute $x=0$ in the given cubic equation, it’ll give $0=1$. Which can’t be true. So there is only one possibility that is true.
So answer of these question is $x = 1{\text{ ,}} \pm \sqrt 2 $
Note:
So, as you know now, for solving such types of questions you have to first find the power of the equation. After that simplify the given equation and find the roots of the given equation. Whenever you have to solve such a type of question, always try to follow the steps.
Complete step by step answer:
For solving such types of questions you first have to find the power of a given equation. If the power of any equation is n then that equation has n roots.
Here, the power of this equation is $3$. So these equations have $3$ roots.
Now, we have to find the $3$ roots of these equations for solving this question.
So for that, we will simplify these equations.
For that first, we will divide and multiply these equations on both sides by x.
So after dividing and multiplying this equations both side by x, we get,
$ \Rightarrow x({x^2} - 1) = 1$
Now, suppose if $a \times b = 1$ then, there are two possibilities:
$\begin{gathered}
\Rightarrow a = 1{\text{ and }}b = 1 \\
\Rightarrow a = - 1{\text{ and }}b = - 1 \\
\end{gathered} $
So, here $a = x$and $b = {x^2} - 1$
$\begin{gathered}
\Rightarrow x = 1{\text{ and }}{{\text{x}}^2} - 1 = 1 \\
\Rightarrow x = - 1{\text{ and }}{{\text{x}}^2} - 1 = - 1 \\
\end{gathered} $
Now, we will solve of these question first for$x = 1{\text{ }}$;
$\begin{gathered}
\Rightarrow {{\text{x}}^2} - 1 = 1 \\
\Rightarrow {x^2} = 2 \\
\Rightarrow x = \pm \sqrt 2 \\
\end{gathered} $
So, for first possibility x have $3$ roots which is given below:
$x = 1{\text{ ,}} \pm \sqrt 2 $
Now, for second possibility:
If $x = - 1{\text{ }}$
Then we get,
$
\Rightarrow {{\text{x}}^2} - 1 = - 1 \\
\Rightarrow {x^2} = 0 \\
\Rightarrow {\text{not possible}} \\
$
Because when we substitute $x=0$ in the given cubic equation, it’ll give $0=1$. Which can’t be true. So there is only one possibility that is true.
So answer of these question is $x = 1{\text{ ,}} \pm \sqrt 2 $
Note:
So, as you know now, for solving such types of questions you have to first find the power of the equation. After that simplify the given equation and find the roots of the given equation. Whenever you have to solve such a type of question, always try to follow the steps.
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