IF: X² + 1/X² = 48
and X + 1/X = 7
then prove that X³ + 1/X³ = 329
$\displaystyle \begin{align*}{x^3} + \frac{1}{{{x^3}}} &= \left( {x + \frac{1}{x}} \right)\left( {{x^2} - x\left( {\frac{1}{x}} \right) + \frac{1}{{{x^2}}}} \right)\\& = 7\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\\& = 7\left( {48 - 1} \right)\\& = ~? \end{align*}$
neelsvanas,
you see what is in the quote box? Idea pointed out that the solutions for your first equation are not that for your
second equation in post #1.
Your problem is wrong (with a couple of numbers). Here is a correct restatement:
If $\displaystyle \ x^2 + \dfrac{1}{x^2} \ = \ $ 47
and $\displaystyle \ x + \dfrac{1}{x} \ = \ $ 7,
then prove that $\displaystyle \ x^3 + \dfrac{1}{x^3} \ = \ $ 322.
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A more challenging problem would have been:
If $\displaystyle \ x + \dfrac{1}{x} \ = \ $ 7,
then determine the value of $\displaystyle \ \ x^3 + \dfrac{1}{x^3} $.
Wanna proof? Havva look:
https://www.wolframalpha.com/input/?...+x%2B1%2Fx%3D7