Question

if x^3+1/x^3=110, find the value of x+1/x

Answer 1

Let $x+\dfrac{1}{x}=k.$

Then,

$x^3+\dfrac{1}{x^3}=x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)-3\left(x+\dfrac{1}{x}\right)=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=k^3-3k$

Given that,

$\begin{aligned}&x^3+\dfrac{1}{x^3}=110\\ \\ \Longrightarrow\ \ &k^3-3k=110\\ \\ \Longrightarrow\ \ &k^3-3k-110=0\\ \\ \Longrightarrow\ \ &k^3-5k^2+5k^2-25k+22k-110=0\\ \\ \Longrightarrow\ \ &k^2(k-5)+5k(k-5)+22(k-5)=0\\ \\ \Longrightarrow\ \ &(k-5)(k^2+5k+22)=0\end{aligned}$

From this, we get,

$k=\mathbf{5}\ \ \ \Longrightarrow\ \ \ x+\dfrac{1}{x}=\mathbf{5}$

Two more values for $k=x+\dfrac{1}{x}$ can be got by factorising $k^2+5k+22.$ Let's factorise it!!!

$\begin{aligned}&k^2+5k+22=0\\ \\ \Longrightarrow\ \ &k^2+\dfrac{5+3\iota\sqrt{7}}{2}k+\dfrac{5-3\iota\sqrt{7}}{2}k+22=0\\ \\ \Longrightarrow\ \ &k\left(k+\dfrac{5+3\iota\sqrt{7}}{2}\right)+\dfrac{5-3\iota\sqrt{7}}{2}\left(k+\dfrac{5+3\iota\sqrt{7}}{2}\right)=0\\ \\ \Longrightarrow\ \ &\left(k+\dfrac{5+3\iota\sqrt{7}}{2}\right)\left(k+\dfrac{5-3\iota\sqrt{7}}{2}\right)=0\end{aligned}$

From this, we get two more values.

$x+\dfrac{1}{x}=-\dfrac{5+3\iota\sqrt{7}}{2}\\ \\ \\ x+\dfrac{1}{x}=-\dfrac{5-3\iota\sqrt{7}}{2}=\dfrac{3\iota\sqrt{7}-5}{2}$

Hence 3 values as answers are got. It should be better taking $x+\dfrac{1}{x}=5$ as the answer.