Question:

<p><strong>Given:</strong> <span>sin θ + cos θ = a</span> ... (i)</p><p><span>sin³ θ + cos³ θ = b</span> ... (ii)</p><p><strong>Step 1:</strong> From Eq. (i), squaring both sides:</p><p><span>sin² θ + cos² θ + 2 sin θ cos θ = a²</span></p><p><span>1 + 2 sin θ cos θ = a²</span></p><p><span>sin θ cos θ = \frac{a² - 1}{2}</span> ... (iii)</p><p><strong>Step 2:</strong> From Eq. (ii), using the factorization <span>(sin θ + cos θ)³ = sin³ θ + cos³ θ + 3 sin θ cos θ(sin θ + cos θ)</span>:</p><p><span>a³ = b + 3 sin θ cos θ \cdot a</span></p><p><span>a³ = b + 3 \cdot \frac{a² - 1}{2} \cdot a</span></p><p><span>a³ = b + \frac{3a³ - 3a}{2}</span></p><p><strong>Step 3:</strong> Simplifying:</p><p><span>2a³ = 2b + 3a³ - 3a</span></p><p><span>a³ + 2b - 3a = 0</span> ... (iv)</p><p><strong>Step 4:</strong> Comparing with <span>λa + μb + νa = 0</span> (which should be <span>λa + μb + νa = 0</span>):</p><p><span>λ = 1, μ = 2, ν = -3</span></p><p><strong>Step 5:</strong> Since <span>λ + μ + ν = 1 + 2 - 3 = 0</span>, we use the identity:</p><p><span>λ³ + μ³ + ν³ - 3λμν = (λ + μ + ν)(λ² + μ² + ν² - λμ - μν - νλ)</span></p><p>When <span>λ + μ + ν = 0</span>: <span>λ³ + μ³ + ν³ = 3λμν</span></p><p><span>λ³ + μ³ + ν³ = 3(1)(2)(-3) = -18</span></p><p><strong>∴ Answer is (b) -18</strong></p>