Complex Numbers
Cube Roots of Unity
Grade 11
Question:
<p>If <span class="math">x = a + b</span>, <span class="math">y = a\alpha + b\beta</span> and <span class="math">z = a\beta + b\alpha</span>, where <span class="math">\alpha</span> and <span class="math">\beta</span> are complex cube roots of unity, then <span class="math">xyz</span> is equal to</p>
<p>(a) <span class="math">a^2 + b^2</span></p>
<p>(b) <span class="math">a^3 + b^3</span></p>
<p>(c) <span class="math">a^3 b^3</span></p>
<p>(d) <span class="math">a^3 - b^3</span></p>
Step-by-Step Solution
Key Concept: Use the properties of complex cube roots of unity (1 + ω + ω² = 0 and ω³ = 1) to simplify the product xyz systematically. Recognize that α and β are the non-real cube roots of unity.
<p><strong>Step 1: Identify the cube roots of unity.</strong> Let ω be a primitive cube root of unity, so ω³ = 1 and 1 + ω + ω² = 0. We have α and β as the two non-real cube roots: α = ω and β = ω² (or vice versa).</p><p><strong>Step 2: Express the given values.</strong> Given: x = a + b, y = aα + bβ, z = aβ + bα. Substituting α = ω and β = ω²:</p><p>x = a + b</p><p>y = aω + bω²</p><p>z = aω² + bω</p><p><strong>Step 3: Calculate the product xyz.</strong> </p><p>xyz = (a + b)(aω + bω²)(aω² + bω)</p><p><strong>Step 4: Expand (aω + bω²)(aω² + bω).</strong></p><p>(aω + bω²)(aω² + bω) = a²ω·ω² + abω·ω + abω²·ω² + b²ω²·ω</p><p>= a²ω³ + abω² + abω⁴ + b²ω³</p><p>= a²(1) + abω² + ab(ω) + b²(1)</p><p>= a² + b² + abω + abω²</p><p>= a² + b² + ab(ω + ω²)</p><p><strong>Step 5: Use the property 1 + ω + ω² = 0.</strong> This gives ω + ω² = -1.</p><p>= a² + b² + ab(-1)</p><p>= a² + b² - ab</p><p><strong>Step 6: Multiply by (a + b).</strong></p><p>xyz = (a + b)(a² + b² - ab)</p><p>= a(a² + b² - ab) + b(a² + b² - ab)</p><p>= a³ + ab² - a²b + a²b + b³ - ab²</p><p>= a³ + b³</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B