If a = 26 and b = 22, then the value of \({{a^3 \ - \ b^{3}} \over a^2 \ - \ b^{2}}\) – \({{3ab} \over a \ + \ b}\) is ______.

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 12 Dec 2022 Shift 1)

Answer 1

Given:

a = 26 and b = 22

Concept used:

(a² - b²) = (a + b)(a - b)

(a - b)² = a² + b² - 2ab

(a3 - b3) = (a - b)(a² + b² + ab)

Calculation:

\({{a^3 \ - \ b^{3}} \over a^2 \ - \ b^{2}}-{{3ab} \over a \ + \ b}\)

⇒ \({{a^3 \ - \ b^{3}} \over (a \ + \ b)(a\,-\,b)}-{{3ab} \over a \ + \ b}\)

⇒ \({{(a \ - \ b)(a^2+b^2+ab)} \over (a \ + \ b)(a\,-\,b)}-{{3ab} \over a \ + \ b}\)

⇒ \({{(a^2+b^2+ab)} \over a \ + \ b}-{{3ab} \over a \ + \ b}\)

⇒ \({{a^2+b^2+ab-3ab} \over a \ + \ b}\)

⇒ \({{a^2+b^2-2ab} \over a \ + \ b}\)

⇒ \({{(a-b)^2} \over a \ + \ b}\)

⇒ \({{(26-22)^2} \over 26 \ + \ 22}\)

⇒ \({{4^2} \over 48}\)

⇒ \({{16} \over 48}\)

⇒ \({{1} \over 3}\)

∴ The required answer is \({{1} \over 3}\).

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