Question

Consider the given expression ${{x}^{2}}+1=0$ and find the value of ${{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1$.
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: First we will find the roots of the given equation and will get the roots as iota(i) and –(iota)i. Now we will use the property of iota(i) which is as follows:

& {{i}^{4n}}=1 \\\ & {{i}^{4n+1}}=i \\\ & {{i}^{4n+2}}=-1 \\\ & {{i}^{4n+3}}=-i \\\ \end{aligned}$$ _Complete step-by-step answer:_ We have been given $${{x}^{2}}+1=0$$ and asked to find the value of $${{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1$$. First of all we will find the root of the equation $${{x}^{2}}+1=0$$. $$\begin{aligned} & \Rightarrow {{x}^{2}}=-1 \\\ & \Rightarrow x=\pm \sqrt{-1} \\\ & \Rightarrow x=i,x=-i \\\ \end{aligned}$$ Case 1: when x=i $${{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1={{i}^{18}}+{{i}^{12}}+{{i}^{6}}+1$$ Since we know that $${{i}^{4n+2}}=-1$$, $$\Rightarrow {{i}^{18}}={{i}^{\left( 4\times 4+2 \right)}}=-1$$ Also, $$\begin{aligned} & {{i}^{4n}}=1 \\\ & \Rightarrow {{i}^{12}}={{i}^{\left( 4\times 3 \right)}}=1 \\\ \end{aligned}$$ Again, $$\begin{aligned} & {{i}^{4n+2}}=-1 \\\ & \Rightarrow {{i}^{6}}={{i}^{\left( 4\times 1+2 \right)}}=-1 \\\ \end{aligned}$$ Substituting these values we get as follows: $${{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1=-1+1-1+1=0$$ Case 2: when x=-i $$\begin{aligned} & {{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1={{\left( -i \right)}^{18}}+{{\left( -i \right)}^{12}}+{{\left( -i \right)}^{6}}+1 \\\ & ={{\left( -1 \right)}^{18}}.{{i}^{18}}+{{\left( -1 \right)}^{12}}.{{i}^{12}}+{{\left( -1 \right)}^{6}}.{{i}^{6}}+1 \\\ \end{aligned}$$ Since we know that even power of any negative number gives positive value $$\Rightarrow {{\left( -1 \right)}^{18}}=1,{{\left( -1 \right)}^{12}}=1,{{\left( -1 \right)}^{6}}=1$$ $$\Rightarrow {{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1={{i}^{18}}+{{i}^{12}}+{{i}^{6}}+1$$ We have already found the values of $${{i}^{18}},{{i}^{12}},{{i}^{6}}$$. So by substituting these values in the above expression, we get as follows: $$\Rightarrow {{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1=-1+1-1+1=0$$ Hence we got the value of the expression equal to zero in both the cases. Therefore, the correct answer of the question is option A. Note: We can also solve the question without finding the roots of the equation. Since we have been given $${{x}^{2}}=-1$$ we can solve the question by finding the values of $${{x}^{18}},{{x}^{12}},{{x}^{6}}$$. Also be careful while doing calculation and use the property of iota(i) very carefully as there is a chance of a sign mistake.