Should I solve it first?

Thanks Wendy Natuibs , you are right. I didn't see it earlier.

$\begin{aligned} \left(x+\frac{1}{x}\right)^3 & = x^3+3x+\frac{3}{x} +\frac{1}{x^3} \\ \Rightarrow x^3 +\frac{1}{x^3} & = \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right) \\ & = 7^3 - 3(7) = \boxed{322} \end{aligned}$

$\color{#3D99F6}{\text{Previous Solution}}$

Using the identities below and substituting $a = x$ and $b = \frac{1}{x}$ , we have:

$\begin{aligned} a^2+b^2 & = (a+b)^2 - 2ab \\ a^3+b^3 & = (a+b)(a^2+b^2 - ab) \\ \Rightarrow x^2+\frac{1}{x^2} & = \left(x+\frac{1}{x} \right)^2 - 2x\left(\frac{1}{x} \right) = 7^2 - 2 = 47 \\ \Rightarrow x^3 + \frac{1}{x^3} & = \left(x+\frac{1}{x}\right) \left(x^2+\frac{1}{x^2} - 1 \right) = 7(47-1) = \boxed{322} \end{aligned}$

$x + \frac 1x = 7\\{x}^3 + \left(\frac {1}{x}\right)^3 = ?\\let\;x = a\;and\;\frac {1}{x} = b\\so\;(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3\\(a + b)^3 = a^3 + b^3 + 3(a^2 b + ab^2)\\ \left(x + \frac {1}{x}\right)^3 = x^3 + \left(\frac {1}{x}\right)^3 + 3(x^2 \frac 1x + x \left(\frac {1}{x}\right)^2)\\(7)^3 = x^3 + \left(\frac {1}{x}\right)^3 + 3((x + \frac 1x ))\\(7)^3 = x^3 + \left(\frac {1}{x}\right)^3 + 3(7)\\ (7)^3 - 3(7) = x^3 + \left(\frac {1}{x}\right)^3\\ 343 - 21 = x^3 + \left(\frac {1}{x}\right)^3\\x^3 + \left(\frac {1}{x}\right)^3 = 322$

First, ( x + 1/ x )^3 = x^3 + 1/ x^3 * x * 1/x ( x+ 1/x ),
( x + 1/ x )^3 = x^3 + 1/ x^3 + 3( x + 1/x ),

Putting x + 1/x = 7, we get,
7^3 = x^3 + 1/ x^3 + 3*7,
343 = x^3 + 1/x ^3 + 21,
= x^3 + 1/x ^3 = 343 -21,
x^3 + 1/x^ 3 = 322.

So, the answer is 322.

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$(x+\frac{1}{x})^{2}=7^{2} \\ x^{2}+2\frac{x}{x}+\frac{1}{x^{2}}=49 \\ =>x^{2}+\frac{1}{x^{2}}=47........(1)$

$x^{3}+(\frac{1}{x})^{3}=(x+\frac{1}{x})(x^{2}+\frac{1}{x^{2}}-\frac{x}{x}) \\ =(7)(47-1)=\boxed{322}$

x + 1/x = 7

x^3 + 1/x^3 = x^3 + 1/x^3 + 3(x + 1/x)

= x^3 + 1/x^3 + 3(7)

7^3 = x^3 + 1/x^3 + 21

x^3 + 1/x^3 = 343 - 21

= 322

[extra spaces for clarification]

$x + \frac{1}{x} = 7$ $(x + \frac{1}{x} )^{3} = 7^{3}$ $(x + \frac{1}{x} )^{2}(x + \frac{1}{x}) = 343$ $(x^{2} + 2 + \frac{1}{x^{2}})(x + \frac{1}{x}) = 343$ $x(x^{2} + 2 + \frac{1}{x^{2}}) + \frac{1}{x}(x^{2} + 2 + \frac{1}{x^{2}}) = 343$ $x^{3} + 2x + \frac{1}{x} + x + \frac{2}{x} + \frac{1}{x^{3}} = 343$ $x^{3} + 3x + \frac{3}{x} + \frac{1}{x^{3}} = 343$ $x^{3} + 3 (x + \frac{1}{x} ) + \frac{1}{x^{3}} = 343$ $x^{3} + 3 (7 ) + \frac{1}{x^{3}} = 343$ $x^{3} + 21 + \frac{1}{x^{3}} = 343$ $\boxed{x^{3} + \frac{1}{x^{3}} = 322}$

(x + 1/x )^3 = 343 also it is equal to x^3 + 1/x^3 + 3 x (1/x)(x+ 1/x) by substituting the values x^3 + 1/x^3 + 21 = 343 it implies that x^3 + 1/x^3 = 343-21 = 322

$a^3 + b^3 = (a+b)^3 -3ab(a+b)$ Substitute $a = x, b =\frac {1}{x}$ We get, $x^3 + \frac{1}{x^3} = (x +\frac{1}{x})^3 - 3(x + \frac {1}{x})$ $= (7)^3 - 3(7) = 343 - 21 = 322$

Multiply the first equation by x to get a quadratic solve for x. x=6.8541. Place the value of x into lower equation =322

Cube it minus 3 times itself you get answer.

x+1/x=7
(x+1/x)^2=49
x^2+1/x^2=47
(x^2+1/x^2)(x+1/x)=(47)(7)
x^3+1/x^3+x+1/x=329
x^3+1/x^3=322

While doing this, I also noticed that you could just do whatever x+1/x is equal to (in this case 7) and put it to the 3rd power because that's the highest degree of what you're trying to find (from x^3+1/x^3). Then you can subtract the highest degree times what x+1/x is equal to from it. So in other words, 7^3 - (3)(7). I don't think it works for all of these types of problems though, but I tried some that had 3 as the highest degree of what you're asked to find, and it seemed to work.

x+ 1/x = 7 => (x + 1/x)^2 = 49 => x^2 + 2 + 1/x^2 = 49 => x^2 + 1/x^2 = 47

x^3 + 1/x^3 = (x + 1/x)(x^2 - 1 + 1/x^2) = 7 * (47 -1) = 7 * 46 = 322