Given the quadratic equation 3x2 − 4x + 2 = 0 with roots alpha and beta, show that alpha3 + beta3 = -8/27.

The given quadratic equation is $3x^2 - 4x + 2 = 0$.
For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha \beta = \frac{c}{a}$.
Here, $a=3$, $b=-4$, $c=2$.
Sum of roots:

$$\alpha + \beta = -\frac{-4}{3} = \frac{4}{3}$$

Product of roots:

$$\alpha \beta = \frac{2}{3}$$

The identity for $\alpha^3 + \beta^3$ is $(\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.
Substitute the values of $\alpha + \beta$ and $\alpha \beta$:

$$\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)^3 - 3\left(\frac{2}{3}\right)\left(\frac{4}{3}\right)$$ $$\alpha^3 + \beta^3 = \frac{4^3}{3^3} - \frac{3 \times 2 \times 4}{3 \times 3}$$ $$\alpha^3 + \beta^3 = \frac{64}{27} - \frac{24}{9}$$

To subtract the fractions, find a common denominator, which is 27.

$$\alpha^3 + \beta^3 = \frac{64}{27} - \frac{24 \times 3}{9 \times 3}$$ $$\alpha^3 + \beta^3 = \frac{64}{27} - \frac{72}{27}$$ $$\alpha^3 + \beta^3 = \frac{64 - 72}{27}$$ $$\alpha^3 + \beta^3 = -\frac{8}{27}$$

This shows that $\boxed{\alpha^3 + \beta^3 = -\frac{8}{27}}$.

Let the new roots be $\gamma = \frac{1}{\alpha^3}$ and $\delta = \frac{1}{\beta^3}$.
The sum of the new roots is:

$$\gamma + \delta = \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\beta^3 + \alpha^3}{\alpha^3 \beta^3}$$

From part (a), we know $\alpha^3 + \beta^3 = -\frac{8}{27}$.
The product $\alpha^3 \beta^3$ can be written as $(\alpha \beta)^3$.
We know $\alpha \beta = \frac{2}{3}$.
So, $(\alpha \beta)^3 = \left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$.
Now substitute these values into the sum of new roots:

$$\gamma + \delta = \frac{-\frac{8}{27}}{\frac{8}{27}} = -1$$

The product of the new roots is:

$$\gamma \delta = \left(\frac{1}{\alpha^3}\right) \left(\frac{1}{\beta^3}\right) = \frac{1}{\alpha^3 \beta^3} = \frac{1}{(\alpha \beta)^3}$$

Using $(\alpha \beta)^3 = \frac{8}{27}$:

$$\gamma \delta = \frac{1}{\frac{8}{27}} = \frac{27}{8}$$

A quadratic equation with roots $\gamma$ and $\delta$ is given by $x^2 - (\gamma + \delta)x + \gamma \delta = 0$.
Substitute the sum and product of the new roots:

$$x^2 - (-1)x + \frac{27}{8} = 0$$ $$x^2 + x + \frac{27}{8} = 0$$

To obtain integral coefficients, multiply the entire equation by 8:

$$8\left(x^2 + x + \frac{27}{8}\right) = 8(0)$$ $$8x^2 + 8x + 27 = 0$$

The equation with integral coefficients is $\boxed{8x^2 + 8x + 27 = 0}$.