Given theta + theta = a and 3 theta + 3 theta = b, express b in terms of a.

We are given the equation $\sin \theta + \cos \theta = a$.
Square both sides of the equation:

$$(\sin \theta + \cos \theta)^2 = a^2$$

Expand the left side:

$$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = a^2$$

Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:

$$1 + 2 \sin \theta \cos \theta = a^2$$

Solve for $2 \sin \theta \cos \theta$:

$$2 \sin \theta \cos \theta = a^2 - 1$$

So,

$$\sin \theta \cos \theta = \frac{a^2 - 1}{2}$$

We are given the equation $\sin^3 \theta + \cos^3 \theta = b$.
Use the algebraic identity $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$. Let $x = \sin \theta$ and $y = \cos \theta$:

$$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)$$

Rearrange the terms inside the second parenthesis:

$$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)((\sin^2 \theta + \cos^2 \theta) - \sin \theta \cos \theta)$$

Substitute $\sin^2 \theta + \cos^2 \theta = 1$:

$$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta)$$

We have $\sin \theta + \cos \theta = a$ and $\sin \theta \cos \theta = \frac{a^2 - 1}{2}$. Substitute these into the equation from Step 2:

$$b = (a)\left(1 - \frac{a^2 - 1}{2}\right)$$

Simplify the expression inside the parenthesis:

$$b = a\left(\frac{2}{2} - \frac{a^2 - 1}{2}\right)$$ $$b = a\left(\frac{2 - (a^2 - 1)}{2}\right)$$ $$b = a\left(\frac{2 - a^2 + 1}{2}\right)$$ $$b = a\left(\frac{3 - a^2}{2}\right)$$

Multiply both sides by 2:

$$2b = a(3 - a^2)$$

Distribute $a$ on the right side:

$$2b = 3a - a^3$$

Rearrange the terms to set the equation to zero:

$$a^3 - 3a + 2b = 0$$

This equation eliminates $\theta$.