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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 ?

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    Absolutely! I would be happy changing x=y+z to y=x-z in 7th grade!2012-07-02

5 Answers 5

1

$x=\frac 1 {(5-x)}$

so $x^3 + \frac 1 {x^3} = x^3 + (5-x)^3$

= (cubic term cancels) $125 - 75x +15x^2$

= $125 -15x(5-x) = 125-15 = 110$

7

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.

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    Nice answer. I would like to +1 this another time of writing $(x+y)(x+y)$ instead of $(x+y)^2$, hence avoiding some one into believing the common blooper $(x+y)^2=x^2+y^2$. Also, these identities hold in a field of characteristic two as well.2012-03-16
6

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 $x \neq 0$).

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    I see. You have divided the two polynomials and swept all the calculations under the carpet. : )2012-03-15
5

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 $23$

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    Yep so the assumption I was looking for is already in the question I suppose2012-03-16
3

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 $