If $a$ is a root of the equation $x^{2} - 3x + 1 = 0$, then the value of $\dfrac {a^{3}}{a^{6} + 1}$ is

Since a is a root of x^2 - 3x + 1 = 0, it satisfies a^2 = 3a - 1. Cubing this relationship gives a^6 = (3a - 1)^3 = 27a^3 - 27a^2 + 9a - 1, and substituting a^2 again gives a^6 = 27a^3 - 27(3a - 1) + 9a - 1 = 27a^3 - 72a + 26. The expression a^6 + 1 simplifies to 27a^3 - 72a + 27, and dividing this by a^3 gives (a^6 + 1) / a^3 = 27 - 72/a^2 + 27/a^3. Solving for a^3 / (a^6 + 1) directly by substituting a^3 = a(3a - 1) = 3a^2 - a = 3(3a - 1) - a = 8a - 3 and a^6 = (8a - 3)^2 = 64a^2 - 48a + 9 = 144a - 64 + 9 = 144a - 55, we find a^6 + 1 = 144a - 54, which does not easily yield 1/18. Using the standard manipulation, divide x^2 - 3x + 1 = 0 by x to get x - 3 + 1/x = 0, so x + 1/x = 3; cubing both sides gives x^3 + 1/x^3 = 27 - 9 = 18. Therefore, a^3 / (a^6 + 1) is equivalent to 1 / (a^3 + 1/a^3) = 1/18.

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