If \(x+\dfrac{1}{x}=1\) then evaluate \(x^{29}+\dfrac{1}{x^{89}}\) [solved]

Question

If \(x+\dfrac{1}{x}=1\) then evaluate \(x^{29}+\dfrac{1}{x^{89}}\)

Answer 1

\(1\)

 

Explanation:

\(x+\frac{1}{x}=1\)

\(\Rightarrow \quad x^{2}+1=x\)

\(\Rightarrow x^{2}-x+1=0\)

\(\Rightarrow(x-1)\left(x^{2}-x+1\right)=0\)

\(\Rightarrow x^{3}-1=0\)

\(\Rightarrow x^{3}=1 .\)

 

Given the expression:
\[
x^{29}+\dfrac{1}{x^{89}}
\]

\(=\dfrac{x^{30}}{x}+\dfrac{x}{x^{90}}\)

\(=\dfrac{\left(x^{3}\right)^{10}}{x}+\dfrac{x}{\left(x^{3}\right)^{30}}\)

\(=\dfrac{1}{x}+\frac{x}{1}\)

\(=\dfrac{1}{x}+x=1\)