Prove the equation without solving for X My niece asked me this - If $x=1/(5-x)$ prove $x^3 + \dfrac{1}{x^3}=110$ without solving for x. I said its not possible since solving for x itself gives me two roots for x (one being $\approx4.79$) and substituting for it in the second equation approx gives me 110. So proving algebraically without any assumptions is not possible. Is this right ?
 A: You are given $x = 1/(5-x)$, i.e. $x$ is a root of the polynomial $x^2-5x+1$. Then it is also a root of $$(x^2-5x+1)(x^4+5x^3+24x^2+5x+1) = x^6-110x^3+1,$$
so $x^3+1/x^3 = 110$ (since $x \neq 0$).
A: Hint $\ $ Exploit innate symmetry. For $\rm\:y = x^{-1}\:$ you know $\rm\:xy = 1\:$ and are given $\rm\:x+y = 5\:$ so 
$$\rm  x^2 + y^2\ =\ (x+y)\ (\:x\ +\ y)\  -\  xy\ (1 + 1)\ =\ 5\:\cdot\: 5 - 1\cdot 2\: =\: 23$$
$$\rm\ \  x^3 +  y^3\ =\ (x+y)\ (x^2+y^2) -\: xy\ (x+y)\ =\ 5\cdot 23 - 1\cdot 5\: =\: 110$$
$$\rm\quad\ \: x^{n+1}+y^{n+1}\ =\ (x+y)\ (x^n+y^n) -\ xy\: (x^{n-1}+y^{n-1})\quad for\ \  all\ \ \ n \ge 0\qquad\quad $$ 
Above is a special case of Newton's identities for expressing power sums in terms of elementary summetric polynomials.
A: Well, not really. 

It is quite possible that, both the values of $x$ give the same value of $E=x^3+\dfrac 1 {x^3}$ and hence you might be able to get that without solving for $x$'s. 

(If you are not convinced that two different $x$ could give you same value of $E$, think what would happen if $k$ and $\dfrac 1 k$ were two values of $x$, you got.) (This is actually the case here.)


You have that  $5-x=\dfrac 1 x \implies x+\dfrac 1 x=5$
Hint to this problem: What is $\left(x+\dfrac 1 x\right)^3$?

  Now $x^3+3x+3\dfrac 1 x+\dfrac 1 {x^3}=125 \implies x^3+\dfrac 1 {x^3}=110 $


Additional Exercise:
What is $x^2+ \dfrac{1}{x^2}$? 

Hint: Consider $\left(x+ \dfrac 1 x \right)^2$. Answer is $23$

A: From your equation you get $5x-x^2=1$, so $5=\frac{x^2+1}{x}$. Now remind the following cubic fromula $(a+b)^3=a^3+b^3+3ab(a+b)$, then
$$
5^3=\left(\frac{x^2+1}{x}\right)^3=\left(x+x^{-1}\right)^3=x^3+x^{-3}+3xx^{-1}(x+x^{-1})=
$$
$$
x^3+x^{-3}+3\frac{x^2+1}{x}=x^3+x^{-3}+15.
$$
So you get
$$
x^3+x^{-3}=125-15=110
$$
A: $x=\frac 1 {(5-x)}$
so $x^3 + \frac 1 {x^3} = x^3 + (5-x)^3$
= (cubic term cancels) $125 - 75x +15x^2$
= $125 -15x(5-x) = 125-15 = 110$
