If \(x+\dfrac{1}{x}=2\sqrt{3}\), then find the value of \(x^3+\dfrac{1}{x^3}\).
\(18\sqrt{3}\)
To find the value of \(x^3+\dfrac{1}{x^3}\) given that \(x+\dfrac{1}{x}=2\sqrt{3}\), we can utilize algebraic identities. Let's solve this problem step by step.
- Start with the identity for cubes: \(x^3 + \dfrac{1}{x^3} = \left(x + \dfrac{1}{x}\right)^3 - 3\left(x + \dfrac{1}{x}\right)\).
- Substitute the given value: \(x + \dfrac{1}{x} = 2\sqrt{3}\).
- Calculate \((x + \dfrac{1}{x})^3\):
- \((2\sqrt{3})^3 = 8 \times 3^{3/2} = 8 \times 3\sqrt{3} = 24\sqrt{3}\).
- Find \(3 \left(x + \dfrac{1}{x}\right)\):
- \(3 \times 2\sqrt{3} = 6\sqrt{3}\).
- Substitute into the identity: \(x^3 + \dfrac{1}{x^3} = 24\sqrt{3} - 6\sqrt{3}\).
- Simplify: \(x^3 + \dfrac{1}{x^3} = 18\sqrt{3}\).
Thus, the value of \(x^3+\dfrac{1}{x^3}\) is \(18\sqrt{3}\), which corresponds to the correct option.
Similar Questions
- Given, $x = \sqrt{3}$, what is $(x + 1)^2 + (x - 1)^2$?
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