If x = 2 - \(2^\frac{1}{3}\)+ \(2^\frac{2}{3}\), then find the value of x³ - 6x² + 18x.

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 13 Dec 2022 Shift 4)

Answer 1

Given:

x = 2 - \(2^\frac{1}{3}\) + \(2^\frac{2}{3}\)

Formula used:

(a - b)³ = a³ - b³ - 3ab (a - b)

Calculation:

⇒ x = 2 - \(2^\frac{1}{3}\) + \(2^\frac{2}{3}\)

⇒ x - 2 = \(2^\frac{2}{3}\) - \(2^\frac{1}{3}\) -------- (1)

Cubing both sides

⇒ (x - 2)³ = (\(2^\frac{2}{3}\) - \(2^\frac{1}{3}\))³

⇒ x³ - 8 - 3 × 2 × x (x - 2) = 2² - 2 - 3 × \(2^\frac{2}{3}\) × \(2^\frac{1}{3}\) (\(2^\frac{2}{3}\) - \(2^\frac{1}{3}\))

⇒ x3 - 8 - 6x² + 12x = 2 - 6 ×( x - 2) [from eq. (1)]

⇒ x3 - 8 - 6x2 + 12x = 2 - 6x + 12

⇒ x3 - 6x2 + 18x = 14 + 8 = 22

∴ The correct answer is 22.

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