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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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It is impressive to me that this is 7th grade math now. Just saying. –  Alex Hirzel Jul 1 '12 at 18:05
    
Absolutely! I would be happy changing x=y+z to y=x-z in 7th grade! –  Slartibartfast Jul 2 '12 at 5:56

5 Answers 5

up vote 5 down vote accepted

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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My 7th grade niece will be very happy with this explanation(And well exactly how I solved it later). Thank you.But still 7th grade math makes me think different now :D x is approximately 4.79 but x+1/x is assumed to be rounded to 5 from 4.99 from the given equation which is clever. –  Slartibartfast Mar 16 '12 at 5:44
    
Yep so the assumption I was looking for is already in the question I suppose –  Slartibartfast Mar 16 '12 at 5:53

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. –  user21436 Mar 16 '12 at 7:48

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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It would be nice if you told us you're considering the polynomial whose roots are cubes of roots of $x^2-5x+1$ and hence the result follows. –  user21436 Mar 15 '12 at 11:50
    
@KannappanSampath: How do you compute the polynomial whose roots are cubes of roots of $x^2-5x+1$? –  marlu Mar 15 '12 at 12:12
    
Well, then what are you doing? (And, that is computable and not hard.) –  user21436 Mar 15 '12 at 12:14
    
I just expressed $x=1/(5-x)$ and $x^3+1/x^3=110$ as polynomial equations and used the fact that "every root of a polynomial $p(x)$ is also a root of $q(x)$" is equivalent to $p$ dividing $q$. –  marlu Mar 15 '12 at 12:20
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$x$ satisfying the equation $x^3+1/x^3 = 110$ is clearly equivalent to $x$ being a root of the polynomial $X^6-110X^3+1$. I don't see what is wrong with that. Similarly, $x=1/(5-x)$ is equivalent to $x$ being a root of the polynomial $X^2-5X+1$. So showing that every solution to $x=1/(5-x)$ satisfies $x^3+1/x^3=110$ is equivalent to showing that each root of $X^2-5X+1$ is also a root of $X^6-110X^3+1$. This, in turn, is the case iff $X^2-5X+1$ divides $X^6-110X^3+1$ as a polynomial, and this is indeed the case. Please let me know if this explanation is wrong or still unclear. –  marlu Mar 15 '12 at 13:45

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 $$ x^3+x^{-3}=125-15=110 $$

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

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

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

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

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