Simultaneous Equations in Three Variables!

$\displaystyle \sum_{\text{cyc}} \tan^{-1}x = \pi \Rightarrow x + y + z = xyz$

$x + y + z = 6$

$xy + yz + xz = 11$

Let us consider a monic cubic polynomial having roots $x,y,z$ .

For the sake of clarity, let us call it $f(u)$

$f(u) = u^3 - 6u^2 + 11u - 6$

Since the sum of coefficients of $f(u)$ is $0$ , $1$ is a root of $f(u)$

$f(u) = (u-1)(u^2 - 5u + 6) = (u-1)(u-2)(u-3)$

$\Rightarrow f(u) = 0 , u = 1,2,3$

Hence the values of $x,y,z$ are $1,2,3$ respectively.

$x^5 + y^5 + z^5 = 1 + 32 + 243 = \boxed{276}$

---

Note:

Why $\displaystyle \sum_{\text{cyc}} \tan^{-1}x = \pi \Rightarrow x + y + z = xyz$ ?

$\pi = \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1} \left(\dfrac{x + y + z - xyz}{1 - xy - yz - xz} \right)$

Taking $\tan$ on both sides, we get

$\tan^{-1} \left(\dfrac{x + y + z - xyz}{1 - xy - yz - xz} \right) = 0$

$\Rightarrow x + y + z - xyz = 0 \Rightarrow x + y + z = xyz$

• Please upvote if you like the solution.