Question

• If x4 - 124x2+1 = 0 then find x3+1/x3?​

Answer 1

Hope this will help u

Answer 2

Solution :-

→ x⁴ - 124x² + 1 = 0

→ x²(x² - 124 + 1/x²) = 0

→ x² + 1/x² = 124

→ x² + 1/x² + 2 = 124 + 2

→ (x)² + (1/x)² + 2 * x * (1/x) = 126

→ (x + 1/x)² = 126

→ (x + 1/x) = ±√126

taking √126 ,

→ (x + 1/x) = √126

→ (x + 1/x)³ = 126√126

→ x³ + 1/x³ + 3 * x * 1/x * (x + 1/x) = 126√126

→ x³ + 1/x³ + 3 * √126 = 126√126

→ (x³ + 1/x³) = 126√126 - 3√126

→ (x³ + 1/x³) = 123√126 (Ans.)

taking (-√126),

→ (x + 1/x) = -√126

→ (x + 1/x)³ = -126√126

→ x³ + 1/x³ + 3 * x * 1/x * (x + 1/x) = 126√126

→ x³ + 1/x³ + 3 * (-√126) = -126√126

→ (x³ + 1/x³) = -126√126 + 3√126

→ (x³ + 1/x³) = (-123√126) (Ans.)

Learn more :-

Let a, b and c be non-zero real numbers satisfying (a³)/(b³ + c³) + (b³)/(c³ + a³) + (c³)/(a³ + b³)

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if a²+ab+b²=25

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