If α, β are roots of x² - x - 1 = 0, find the value of 2α⁴ + β³ — JEE Mathematics

From the equation: $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$.
Let's compute higher powers:

$$\alpha^3 = \alpha(\alpha^2) = \alpha(\alpha + 1) = \alpha^2 + \alpha = (\alpha + 1) + \alpha = 2\alpha + 1$$

$$\alpha^4 = \alpha(\alpha^3) = \alpha(2\alpha + 1) = 2\alpha^2 + \alpha = 2(\alpha + 1) + \alpha = 3\alpha + 2$$

$$\beta^3 = 2\beta + 1$$

Substitute these back into the target expression:

$$2\alpha^4 + \beta^3 = 2(3\alpha + 2) + (2\beta + 1) = 6\alpha + 4 + 2\beta + 1 = 4\alpha + 2(\alpha + \beta) + 5$$

We know from the sum of roots that $\alpha + \beta = 1$:

$$= 4\alpha + 2(1) + 5 = 4\alpha + 7$$

Since the problem specifies a definite numerical relation or reduction, keeping it in terms of $\alpha$ yields $4\alpha + 7$.

Answer: $4\alpha + 7$