One way would be to use the power-reduction trigonometric identity:

$$ \cos^3(x) = \frac{3 \cos(x) + \cos(3x)}{4} $$

Due to the linearity property of the Fourier transform, you can transform each term separately and take their weighted sum to get the transform of the entire expression. The relationship we will use (from line 304 here) is:

$$ \mathcal{F}\{\cos(ax)\} = \pi\left(\delta(\omega - a) + \delta(\omega + a)\right) $$

Which assumes that you're using the non-unitary, angular frequency definition of the Fourier transform:

$$ \mathcal{F}\{x(t)\} = X(\omega) = \int_{-\infty}^{\infty}x(t) e^{-j\omega t}dt $$

This would yield:

$$ \begin{align} \mathcal{F}\{\cos^3(x)\} &= 3 \mathcal{F}\{\cos(x)\} + \mathcal{F}\{\cos(3x)\} \\ &= \frac{\pi}{4}\left(3 \delta(\omega - 1) + 3\delta(\omega + 1) + \delta(\omega - 3) + \delta(\omega + 3) \right) \end{align} $$

One way would be to use the power-reduction trigonometric identity:

$$ \cos^3(x) = \frac{3 \cos(x) + \cos(3x)}{4} $$

Due to the linearity property of the Fourier transform, you can transform each term separately and take their weighted sum to get the transform of the entire expression. The relationship we will use (from line 304 here) is:

$$ \mathcal{F}\{\cos(ax)\} = \pi\left(\delta(\omega - a) + \delta(\omega + a)\right) $$

Which assumes that you're using the non-unitary, angular frequency definition of the Fourier transform:

$$ \mathcal{F}\{x(t)\} = X(\omega) = \int_{-\infty}^{\infty}x(t) e^{-j\omega t}dt $$

This would yield:

$$ \begin{align} \mathcal{F}\{\cos^3(x)\} &= \frac 34 \mathcal{F}\{\cos(x)\} + \frac 14 \mathcal{F}\{\cos(3x)\} \\ &= \frac{\pi}{4}\left(3 \delta(\omega - 1) + 3\delta(\omega + 1) + \delta(\omega - 3) + \delta(\omega + 3) \right) \end{align} $$