I'm a bit stuck: IF: X² + 1/X² = 48 and X + 1/X = 7 then prove that X³ + 1/X³ = 329

Hello, and welcome to FMH! :)

We are told:

[MATH]x+\frac{1}{x}=7[/MATH]
What do you get when you cube both sides?
 
what is the rule when from : X³ + 1/X³ = 329 when the next step is : X + 1/X ( X² + 1/X²)= 329
 
neilvanas said:
ok, then we get X² + 1/X² = 46
Click to expand...

What I meant was to begin with the given:

[MATH]x+\frac{1}{x}=7[/MATH]
Cube both sides:

[MATH]\left(x+\frac{1}{x}\right)^3=7^3[/MATH]
And this can be written, using the binomial theorem on the left:

[MATH]x^3+3x^2\frac{1}{x}+3x\frac{1}{x^2}+\frac{1}{x^3}=343[/MATH]
This simplifies to:

[MATH]x^3+3\left(x+\frac{1}{x}\right)+\frac{1}{x^3}=343[/MATH]
Can you take it on home now? :)
 
neilvanas said:
hi, ok I did, but it equals 322, right?
Click to expand...

Yes, the problem is flawed. If we take this as true:

[MATH]x+\frac{1}{x}=7[/MATH]
Then neither of the other two statements can be true. We must then have:

[MATH]x^2+\frac{1}{x^2}=47[/MATH]
[MATH]x^3+\frac{1}{x^3}=322[/MATH]
 
Last edited:
Whoa .. that's a bit trippy!
I approached it like this:

Using \(\displaystyle a^3 + b^3 = (a+b)(a^2- ab +b^2)\)

So
\(\displaystyle x^3 + \frac{1}{x^3} = (x + \frac{1}{x})(x^2 - 1 + \frac{1}{x^2})\)

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

\(\displaystyle x^3 + \frac{1}{x^3} = 7 * (48 - 1)\)

\(\displaystyle x^3 + \frac{1}{x^3} = 329\)
 
neilvanas said:
IF: X² + 1/X² = 48
and X + 1/X = 7
then prove that X³ + 1/X³ = 329
Click to expand...
\(\displaystyle \begin{align*}{x^3} + \frac{1}{{{x^3}}} &= \left( {x + \frac{1}{x}} \right)\left( {{x^2} - x\left( {\frac{1}{x}} \right) + \frac{1}{{{x^2}}}} \right)\\& = 7\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\\& = 7\left( {48 - 1} \right)\\& = ~? \end{align*}\)
 
But if \(\displaystyle x + \frac {1}{x} =7\)
then
\(\displaystyle (x + \frac{1}{x})^2 = x^2 +2 + \frac{1}{x^2} =49\)
so
\(\displaystyle x^2+\frac{1}{x^2} =47\) not \(\displaystyle 48\)

So yes the question is flawed!

But if you approached it like I had above ( edit: and pka too!) you may not have realised that. Interesting!
 
The problem is like if I said this:

If x = 5 and 2x = 11, prove that 3x = 16.​

I love problems like this that you can solve if you're "creative" enough to see the "trick", but can realize are unsolvable if you don't take the bait. It's a double-cross, tricking the trickster! (And very possibly the poser, as well.)

Shortcuts can be fun but dangerous.
 
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