Question:

Step 1: Apply Vieta's formulas to find the sum and product of roots. For the quadratic equation $3x^2 - x + k = 0$ with roots $\sin\theta$ and $\cos\theta$, Vieta's formulas give us: $$\sin\theta + \cos\theta = \frac{1}{3}$$ $$\sin\theta\cos\theta = \frac{k}{3}$$ Step 2: Use the identity $(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta$ to find $k$. Squaring the sum of roots: $$\left(\sin\theta + \cos\theta\right)^2 = 1 + 2\sin\theta\cos\theta$$ Substituting the known values: $$\left(\frac{1}{3}\right)^2 = 1 + 2 \cdot \frac{k}{3}$$ $$\frac{1}{9} = 1 + \frac{2k}{3}$$ $$\frac{2k}{3} = \frac{1}{9} - 1 = -\frac{8}{9}$$ $$k = -\frac{4}{3}$$ Step 3: Calculate the product of roots using the value of $k$. $$\sin\theta\cos\theta = \frac{k}{3} = \frac{-4/3}{3} = -\frac{4}{9}$$ Step 4: Use the sum of cubes factorization to find $\sin^3\theta + \cos^3\theta$. We use the algebraic identity: $$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta)$$ Since $\sin^2\theta + \cos^2\theta = 1$: $$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$$ Substituting our values: $$\sin^3\theta + \cos^3\theta = \frac{1}{3}\left(1 - \left(-\frac{4}{9}\right)\right)$$ $$= \frac{1}{3}\left(1 + \frac{4}{9}\right) = \frac{1}{3} \cdot \frac{13}{9} = \frac{13}{27}$$ Step 5: Calculate $54(\sin^3\theta + \cos^3\theta)$ to find the final answer. $$54(\sin^3\theta + \cos^3\theta) = 54 \cdot \frac{13}{27} = \frac{54 \cdot 13}{27} = 2 \cdot 13 = 26$$ **Final Answer: 26** The answer is **Option 2: 26**