Question

find the value of (1). x³+8y³-36xy-216 when x=2y+6

Answer 1

Given that, x = 2y + 6
The equation,
x³ + 8y³ - 36xy -216
=> (2y+6)³ - 36(2y+6)y+ 8y³-216
=> 8y³ + 3×4y²×6 +3×2y×36+6y³ -72y² + 216y-216+8y³
=> 8y³ + 72y² + 216y- 216y³-72y²+216y-216 +8y³
(1) => 8 +72 +216 -216 -72 +216 -216 +8
=> 16 [ Answer]

Answer 2

Answer :

Given, x = 2y + 6

Taking cube to both sides, we get

x³ = (2y + 6)³

⇒ x³ = (2y)³ + {3 × (2y)² × 6} + {3 × 2y × 6²} + 6³

⇒ x³ = 8y³ + (3 × 4y² × 6) + (3 × 2y × 36) + 216

⇒ x³ = 8y³ + 72y² + 216y + 216

Now, x³ + 8y³ - 36xy - 216

= (8y³ + 72y² + 216y + 216) + 8y³ - 36(2y + 6)y - 216

= 8y³ + 72y² + 216y + 216 + 8y³ - 72y² - 216y - 216

= (8 + 8)y³ + (72 - 72)y² + (216 - 216)y + (216 - 216)

= 16y³

Identity Rule :

(a + b)³ = a³ + 3a²b + 3ab² + b³

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