Question

find the value of x³ -8y³ - 36xy- 216 when x=2y + 6​

Answer 1

(2y + 6)^3 - 8*y^2 - 36*(2y + 6)*y = 216

8*y^3 - 8*y^2 + 216 = 216

y^3 - y^2 = 0

y - 1 = 0;

y = 1 & 0

Answer 2

Solution:

We have,

x³ - 8y³ - 36xy - 216

⟼ x³ + (-8y³) + (-216) - 36xy

⟼ x³ + (-2y)³ + (-6)³ - 3 × x × (-2y) × (-6)

Formula Used:

a³ + b³ + c³ - 3abc where a = x , b = -2y , c = -6

After substituting the values,

⟼ (a + b + c) (a² + b² + c² - ab - bc - ca)

⟼ (x - 2y - 6) (x² + 4y² + 36 + 2xy - 12y + 6x)

⟼ 0 × (x² + 4y² + 36 + 2xy - 12y + 6x) = 0

where, [ x - 2y - 6 = 0 ]

Hence,

Proved that x³ - 8y³ - 36xy - 216 = 0.

Extra Dose:

• (a + b)³ = a³ + b³ + 3ab (a+b)
• (a - b)³ = a³ - b³ - 3ab (a -b)
• ( a - b)² = a² + b² - 2ab
• (a + b)² = a² + b² + 2ab

Thanks: