## anonymous one year ago if x+1/x=7, what is x^3+1/x^2?

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1. zepdrix

Is it $$\large\rm \frac{x+1}{x}=7$$ or $$\large\rm x+\frac{1}{x}=7$$ ?

2. anonymous

(x) + (1/x) = 7, the second one

3. zepdrix

Fractions are bad. No bueno. So let's multiply through to get rid of the x in the denominator. Multiplying by x gives us:$\large\rm x^2+1=7x$Ok with that step? What should we do next, what do you think? :)

4. zepdrix

There is probably a cool way to relate the first equation to the second one... but I'm not seeing it. So instead what we can do is.. simply solve for the value of x in the first equation, and plug that value into the second equation.

5. anonymous

I do not think that is the way my teacher would like for me to go about it. I have been staring at this problem for a good 35 minutes and I'm stumped.

6. zepdrix

No? Hmm. Ok lemme make sure I'm understanding the second equation then, is it $$\large\rm x^3+\frac{1}{x^2}$$ ?

7. anonymous

Yes it is.

8. Mertsj

$x+\frac{1}{x}=7$ $x=7-\frac{1}{x}$ $\frac{1}{x}=7-x$

9. zepdrix

I was thinking maybe we could do something like this:$\large\rm x+\frac{1}{x}=7$Then,$\large\rm \left(x+\frac{1}{x}\right)^2=49$and$\large\rm x^2+\frac{1}{x}=47$But that doesn't quite get us the third power :( Darn...

10. zepdrix

Woops typo,$\large\rm x^2+\frac{1}{x^2}=47$

11. Mertsj

$x^3+\frac{1}{x^2}=(7-\frac{1}{x})^3+\frac{1}{(7-x)^2}$

12. freckles

hero had a good way on a previous post like this one something about squaring both sides and also cubing both sides of that one equation

13. freckles

$(x+\frac{1}{x})^2=7^2 \\ (x+\frac{1}{x})^3=7^3$ some expanding will be involved

14. freckles

you will get values for both $x^2+\frac{1}{x^2} \text{ and } x^3+\frac{1}{x^3} \\ \text{ then add those you get } \\ x^3+\frac{1}{x^2}+x^2+\frac{1}{x^3} \\ =x^3+\frac{1}{x^2}+\frac{1}{x}(x^3+\frac{1}{x^2}) \\ =(x^3+\frac{1}{x^2})(1+\frac{1}{x}) \\ \text{ so you have } \\ x^3+\frac{1}{x^2}+x^2+\frac{1}{x^3}=(x^3+\frac{1}{x^2})(1+\frac{1}{x}) \\ \text{ solving for } x^3+\frac{1}{x^2} \\ \text{ you have } \\ x^3+\frac{1}{x^2}=\frac{x^3+\frac{1}{x^2}+x^2+\frac{1}{x^3}}{1+\frac{1}{x}}$ and remember you found the numerator above from doing what hero suggested and then you are given the denominator

15. anonymous

seems like there is a more prosaic way of doing this

16. freckles

you know what I think I want to re-express what I said just a bit: $x^3+\frac{1}{x^2}=\frac{(x^3+\frac{1}{x^3})+(x^2+\frac{1}{x^2})}{1+\frac{1}{x}}$ just wanted to group together what we actually will find from the above equations that I mentioned (or I mean hero had mentioned in a previous post)

17. anonymous

first equation tells you $x=\frac{7\pm3\sqrt{5}}{2}$ it is much less fun, but i am sure will get an answer