Finding $x^3 - \frac{1}{x^3}$ from $x^4 + \frac{1}{x^4} = 322$
We are given the equation $x^4 + \frac{1}{x^4} = 322$ and need to find the value of $x^3 - \frac{1}{x^3}$.
Calculate Intermediate Values
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First, find $x^2 + \frac{1}{x^2}$. We know the identity $(a+b)^2 = a^2 + b^2 + 2ab$. Let $a = x^2$ and $b = \frac{1}{x^2}$.
So, $\left(x^2 + \frac{1}{x^2}\right)^2 = \left(x^2\right)^2 + \left(\frac{1}{x^2}\right)^2 + 2(x^2)\left(\frac{1}{x^2}\right) = x^4 + \frac{1}{x^4} + 2$.
Substitute the given value: $\left(x^2 + \frac{1}{x^2}\right)^2 = 322 + 2 = 324$.
Taking the square root of both sides: $x^2 + \frac{1}{x^2} = \sqrt{324} = 18$. We take the positive root as $x^2 + \frac{1}{x^2}$ is always positive for real $x$.
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Next, find $x - \frac{1}{x}$. We use the identity $(a-b)^2 = a^2 + b^2 - 2ab$. Let $a = x$ and $b = \frac{1}{x}$.
So, $\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2(x)\left(\frac{1}{x}\right) = x^2 + \frac{1}{x^2} - 2$.
Substitute the value of $x^2 + \frac{1}{x^2}$: $\left(x - \frac{1}{x}\right)^2 = 18 - 2 = 16$.
Taking the square root: $x - \frac{1}{x} = \pm\sqrt{16} = \pm 4$.
Calculate the Expression $x^3 - \frac{1}{x^3}$
We use the algebraic identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.
Let $a = x$ and $b = \frac{1}{x}$. Then $x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + (x)\left(\frac{1}{x}\right) + \frac{1}{x^2}\right)$.
Simplify the second factor: $x^2 + 1 + \frac{1}{x^2}$.
Substitute the calculated values: $x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(\left(x^2 + \frac{1}{x^2}\right) + 1\right)$.
Using $x^2 + \frac{1}{x^2} = 18$ and $x - \frac{1}{x} = \pm 4$: $x^3 - \frac{1}{x^3} = (\pm 4)(18 + 1) = (\pm 4)(19)$.
This gives $x^3 - \frac{1}{x^3} = \pm 76$. Since 76 is one of the options, we select it.
Final Answer: The final answer is 76.
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