Question

If \[a\] is a root of the equation \[{{x}^{2}}-3x+1=0\] , then the value of \[\dfrac{{{a}^{3}}}{{{a}^{6}}+1}\] ?

Answer 1

Hint: To find the value of the problem, we first need to find the root of the equation either by separation method or by using quadratic formula. Here we use the quadratic formula which is stated as:
 \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
The values of the variables \[a,b,c\] from the above equation is taken as according to \[a{{x}^{2}}+bx+c\] .

Complete step-by-step answer:
To find the value we first write the value of the variable by forming the equation in form of \[a{{x}^{2}}+bx+c=0\] which is equivalent to \[{{x}^{2}}-3x+1=0\] .
The values of the variables are written in the form of \[a=1\] , \[b=-3\] and \[c=1\] .
Placing the values in the formula of the quadratic equation we get the value of the roots as:
 \[\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
 \[\Rightarrow \dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times \left( 1 \right)}}{2\times 1}\]
Simplifying the given equation, we get:
 \[\Rightarrow \dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4}}{2\times 1}\]
 \[\Rightarrow \dfrac{3}{2}\pm \dfrac{\sqrt{5}}{2}\]
 \[\Rightarrow \dfrac{3}{2}+\dfrac{\sqrt{5}}{2},\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\]
Hence, the value of \[a\] can either be \[\dfrac{3}{2}+\dfrac{\sqrt{5}}{2},\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\] .
Placing either of the values in the question and then find the value of the term \[\dfrac{{{a}^{3}}}{{{a}^{6}}+1}\] . Let us take the value of the \[a\] as \[\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\] which can be simplified as:
 \[\Rightarrow \dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\]
 \[\Rightarrow \dfrac{3+\sqrt{5}}{2}\]
We have taken the value of \[\sqrt{5}=2.2\] .
 \[\Rightarrow 2.61\]
Now we have the value of \[a\] in a simplified term as \[2.61\] . Placing the value of the variable in the term \[\dfrac{{{a}^{3}}}{{{a}^{6}}+1}\] , we get the value of the term as:
 \[\Rightarrow \dfrac{{{a}^{3}}}{{{a}^{6}}+1}\]
 \[\Rightarrow \dfrac{{{\left( 2.61 \right)}^{3}}}{{{\left( 2.61 \right)}^{6}}+1}\]
After placing the values in the term we simplify the term into decimal parts as:
 \[\Rightarrow \dfrac{17.57}{317.11}\]
 \[\Rightarrow \dfrac{1}{18.04}\approx \dfrac{1}{18}\]
So, the correct answer is “$\dfrac{1}{18}$”.

Note: Another method to solve the question is to first change the equation from \[x\] terms to \[a\] term as \[{x}^{2}-3x+1=0\] to \[{a}^{2}-3a+1=0\] now simplifying the equation we get the values as:
 \[\Rightarrow {{a}^{2}}-3a+1\]
 \[\Rightarrow a+\dfrac{1}{a}=3\] .
Now the value of the term \[\dfrac{{{a}^{3}}}{{{a}^{6}}+1}\] , if simplified can be converted into \[\dfrac {1}{{{a}^{3}}+\dfrac{1}{{{a}^{3}}}}\] .
With the value of \[a+\dfrac{1}{a}=3\] and replacing \[{{a}^{3}}+\dfrac{1}{{{a}^{3}}}\] by \[\left( a+\dfrac{1}{a} \right)\left( {{a}^{2}}+\dfrac{1}{{{a}^{2}}}+1 \right)\]
 \[ {{\left( a+b \right)}^{3}}=\left( a-b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\] .
We get the value of \[\dfrac{1}{{{a}^{3}}+\dfrac{1}{{{a}^{3}}}}\] as \[\dfrac{1}{\left( a+\dfrac{1}{a} \right)\left( {{a}^{2}}+\dfrac{1}{{{a}^{2}}}+1 \right)}\] with \[a+\dfrac{1}{a}=3\] .
 \[\Rightarrow \dfrac{1}{\left( a+\dfrac{1}{a} \right)\left( {{a}^{2}}+\dfrac{1}{{{a}^{2}}}+1 \right)}\]
 \[\Rightarrow \dfrac{1}{3\left( 9-3 \right)}\]
 \[\Rightarrow \dfrac{1}{18}\]