Question

if x+1/x=7, what is x^3+1/x^2?

Answer 1

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

Answer 2

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

Answer 3

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? :)

Answer 4

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.

Answer 5

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.

Answer 6

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

Answer 7

Yes it is.

Answer 8

\[x+\frac{1}{x}=7\]
\[x=7-\frac{1}{x}\]
\[\frac{1}{x}=7-x\]

Answer 9

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...

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

seems like there is a more prosaic way of doing this

Answer 16

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)

Answer 17

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