If \(\rm \frac{a}{b}+\frac{b}{a}=1, a \neq 0, b \neq 0\), then find the value of \(\rm \frac{a^3+b^3}{3 a b}\).

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Question

This question was previously asked in

SSC CPO 2024 Official Paper-I (Held On: 28 Jun, 2024 Shift 3)

Answer 1

Given:

a ≠ 0, b = ≠ 0

Formula used:

a³ + b³ = (a + b) (a² + b² - ab)

Calculation:

\(\frac{a}{b} + \frac{b}{a}\) = 1

⇒ (a² + b²)/ab = 1

⇒ a² + b² = ab ---(1)

Now,

a³ + b³

⇒(a + b)(a² + b² - ab)

⇒ (a + b)(ab - ab) [∵ a² + b² = ab]

⇒ (a + b) × 0

⇒ 0

So, \(\frac{a^3+b^3}{3ab}\) = \(\frac{0}{3ab}\) = 0

∴ The correct answer is option (4).

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