If one real root of the quadratic equation 81x² + kx + 256 = 0 is the cube of the other root, then a value of k is

Let the roots of the quadratic equation 81x² + kx + 256 = 0 be α and α³.
By Vieta's formulas, the sum of the roots is α + α³ = -k/81 and the product of the roots is α(α³) = α⁴ = 256/81.
Taking the fourth root of both sides, we get α = ±(256/81)^(1/4) = ±4/3.
If α = 4/3, then α³ = 64/27.
The sum of the roots is α + α³ = 4/3 + 64/27 = (36 + 64)/27 = 100/27.
Then -k/81 = 100/27, which gives k = -300.
If α = -4/3, then α³ = -64/27.
The sum of the roots is α + α³ = -4/3 - 64/27 = (-36 - 64)/27 = -100/27.
Then -k/81 = -100/27, which gives k = 300.
However, none of the options are -300 or 300. Let's check the options.
Let's assume there's a mistake in the question or options. Let's verify the given options.
If k=100, 81x²+100x+256=0. The roots are approximately -0.69 and -4.31. -4.31 is not the cube of -0.69.
If k=144, 81x²+144x+256=0. This quadratic equation factors as (9x+16)²=0, which has a repeated root x=-16/9. Therefore, this is not the correct answer either.
There seems to be an error in either the question or the provided options.