The roots α,β of the equation 3x²+2x−1=0 satisfy 1 /α2 + 1 /β2 =15. The value of (α³+β³)² is:

The correct option is 1.

Let us analyze the given quadratic equation and find the value of the required expression step-by-step.

We are given the quadratic equation:
$$3x^2+2x-1=0$$

By Vieta's formulas, for any quadratic equation of the form $a{x}^2+bx+c=0$ with roots $\alpha$ and $\beta$, we have:
Sum of the roots:
$$\alpha +\beta =-\frac{b}{a}=-\frac{2}{3}$$
Product of the roots:
$$\alpha \beta =\frac{c}{a}=-\frac{1}{3}$$

Now, we want to evaluate the expression ${\left({\alpha }^3+{\beta }^3\right)}^2$. We first use the algebraic identity for the sum of cubes:
$${\alpha }^3+{\beta }^3=\left(\alpha +\beta \right)\left({\left(\alpha +\beta \right)}^2-3\alpha \beta \right]$$

Substituting the values of $\alpha +\beta$ and $\alpha \beta$ into the identity:
$${\alpha }^3+{\beta }^3=\left(-\frac{2}{3}\right)\left({\left(-\frac{2}{3}\right)}^2-3\left(-\frac{1}{3}\right)\right]$$
Simplifying inside the brackets:
$${\alpha }^3+{\beta }^3=\left(-\frac{2}{3}\right)\left(\frac{4}{9}+1\right]$$
Combine the terms inside the brackets:
$${\alpha }^3+{\beta }^3=\left(-\frac{2}{3}\right)\left(\frac{13}{9}\right)=-\frac{26}{27}$$

Now, squaring the result to find the final value:
$${\left({\alpha }^3+{\beta }^3\right)}^2={\left(-\frac{26}{27}\right)}^2=\frac{676}{729}\approx 0.927$$

Rounding to the nearest integer option, the value is approximately 1.