<p>The expression \[\left(\sqrt{2x^2+1}+\sqrt{2x^2-1}\right)^6 + \left(\frac{2}{\sqrt{2x^2+1}+\sqrt{2x^2-1}}\right)^6\] is a polynomial of degree</p>
Step-by-Step Solution
Key Concept: Recognize that the two expressions are reciprocals: if α = (√(2x²+1) + √(2x²-1)), then the second term equals (1/α)⁶. When expanding (α + 1/α)⁶ using binomial theorem, all odd-power irrational terms cancel, leaving only a polynomial.
<p><strong>Step 1:</strong> Let α = √(2x²+1) + √(2x²-1). Note that α · (1/α) = 1.</p><p><strong>Step 2:</strong> Verify the reciprocal relationship: (√(2x²+1) + √(2x²-1)) · (√(2x²+1) - √(2x²-1)) = (2x²+1) - (2x²-1) = 2, so 1/α = (√(2x²+1) - √(2x²-1))/2, but more directly: the second term is 2⁶/(α)⁶ = 64/α⁶.</p><p><strong>Step 3:</strong> Actually, let β = √(2x²+1) - √(2x²-1). Then α·β = 2, so β = 2/α. The given expression is α⁶ + (2/α)⁶ = α⁶ + 2⁶/α⁶.</p><p><strong>Step 4:</strong> Expand using binomial theorem: (α² + 1/α²)³. Since α² = (2x²+1) + (2x²-1) + 2√(4x⁴-1) = 4x² + 2√(4x⁴-1), we need the highest degree terms. The dominant terms come from (4x²)³ = 64x⁶.</p><p><strong>Step 5:</strong> When we compute α⁶ + (64/α⁶), all irrational radicals cancel (by symmetry of binomial expansion), leaving a polynomial. The highest power of x is 6.</p><p>∴ Answer: <strong>B (degree 6)</strong></p>
Correct Answer: B