Question

If $y+\dfrac{1}{y}=1$ then the value of ${{y}^{3}}$ is:
(a)2
(b)1
(c)-1
(d)0

Answer 1

Hint: First we will solve the given equation $y+\dfrac{1}{y}=1$ and convert it in the quadratic equation and find the value of y by solving the root of the quadratic equation and then we will find the value of ${{y}^{3}}$.

Complete step-by-step answer:
Let’s first write the quadratic equation,
$\begin{align}
  & {{y}^{2}}+1=y \\
 & {{y}^{2}}-y+1=0 \\
\end{align}$
First we will use sridharacharya formula to find the roots of this quadratic equation,
The formula is:
If the equation is $a{{x}^{2}}+bx+c=0$ , then the formula for finding x is:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Using this we get,
$\begin{align}
  & y=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4}}{2} \\
 & y=\dfrac{1\pm \sqrt{3}i}{2} \\
\end{align}$
As we have found the value of y now we will use this to find the value of ${{y}^{3}}$ from the value of y.
Now we will find ${{y}^{3}}$,
$\begin{align}
  & ={{\left( \dfrac{1\pm \sqrt{3}i}{2} \right)}^{3}} \\
 & =\dfrac{{{\left( 1\pm \sqrt{3}i \right)}^{3}}}{8} \\
\end{align}$
Here we will use the formula that,
$\begin{align}
  & {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right) \\
 & {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) \\
 & {{i}^{2}}=-1 \\
\end{align}$
Now the value of ${{y}^{3}}$ will be,
$\begin{align}
  & =\dfrac{1+{{\left( \sqrt{3}i \right)}^{3}}+3\sqrt{3}i\left( 1+\sqrt{3}i \right)}{8} \\
 & =\dfrac{1-\left( 3\sqrt{3}i \right)+3\sqrt{3}i-9}{8} \\
 & =\dfrac{-8}{8} \\
 & =-1 \\
\end{align}$
Another value of ${{y}^{3}}$ will be,
$\begin{align}
  & =\dfrac{1-{{\left( \sqrt{3}i \right)}^{3}}-3\sqrt{3}i\left( 1-\sqrt{3}i \right)}{8} \\
 & =\dfrac{1-\left( -3\sqrt{3}i \right)-3\sqrt{3}i-9}{8} \\
 & =\dfrac{-8}{8} \\
 & =-1 \\
\end{align}$
Hence we can see from both expressions we get the same value of y.
Hence, the correct option will be (c).

Note: In this question we have to deal with the complex numbers. So, it’s better to have some knowledge about it and some important formulas that we have already given and it should be memorized by the student for future use. The sridharacharya formula that we have used to find the roots of the equation should also be kept in mind as it’s always useful whenever we deal with quadratic equations.