1. prove the following

IF: X² + 1/X² = 48
and X + 1/X = 7

then prove that X³ + 1/X³ = 329

2. Re: prove the following

Do you know how to factorise the sum of two cubes?

3. Re: prove the following

Originally Posted by neelsvanas
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*}

4. Re: prove the following

the given system of equations is inconsistent

$\displaystyle 49=7^2=\left(x+\frac{1}{x}\right)^2=2+\frac{1}{x^2 }+x^2=50$

5. Re: prove the following

Originally Posted by Idea
the given system of equations is inconsistent

$\displaystyle 49=7^2=\left(x+\frac{1}{x}\right)^2=2+\frac{1}{x^2 }+x^2=50$
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 $\displaystyle \ \ x^3 + \dfrac{1}{x^3}$.