Question

If p = 378, q = 379 and r = 380; then find the value of p³ + q³ + r³ – 3pqr.
1. 3411
2. 3412
3. 3400
4. 3413

Answer 1

Correct Answer - Option 1 : 3411

Given:

p = 378

q = 379

r = 380

Formula Used:

a³ + b³ + c³ – 3abc = 1/2 × (a + b + c) {(a – b)² + (b – c)² + (c – a)²}

Calculation:

p³ + q³ + r³ – 3pqr = 1/2 × (p + q + r) {(p – q)² + (q – r)² + (r – p)²}

⇒ 1/2 × (378 + 379 + 380) × {(378 – 379)² + (379 – 380)² + (380 – 378)²}

⇒ 1/2 × 1137 × 6 = 3411

∴ the required value is 3411

Shortcut trick:

Concept used:

If a, b, c three consecutive numbers, then a³ + b³ + c³ – 3abc = Sum of three numbers × 3

Calculation:

Here p = 378, q = 379, r = 380 i.e, three are consecutive numbers,

⇒ p³ + q³ + r³ – 3pqr = (378 + 379 + 380) × 3 = 1137 × 3 = 3411

∴ the required value is 3411