Question

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Q22. If $\sf{\bigg(x+\dfrac{1}{x}\bigg)=4}$ and $\sf{\bigg(x^2+\dfrac{1}{x^3}\bigg)=12}$, then what would be the value of $\sf{\bigg(x^3+\dfrac{1}{x^2}\bigg)}$?

(1) 20
(2) 26
(3) 66
(4) 54​

Answer 1

OPTION 4 → 54

Step-by-step explanation:

GIVEN

• x + 1/x = 4 _(i)
• x² + 1/x³ = 12

Squaring eqn (i)

• (x + 1/x)² = 16
• x² + 1/x² = 14 _(ii)

Cubing eqn (i)

• (x + 1/x)³ = 4³
• x³ + 1/x³ + 3(x + 1/x) = 64
• x³ + 1/x³ = 52 _(iii)

(ii) + (iii)

• x² + 1/x² + x³ + 1/x³ = 14 + 52
• (x² + 1/x³) + (x³ + 1/x²) = 66
• 12 + x³ + 1/x² = 66
• x³ + 1/x² = 54