Question

Find the value of x3-8y3-36xy-216, at x = 2y+6

Answer 1

so given

x = 2y+6

⇒ x - 2y = 6      ...... (i)

cubing both sides

⇒ x³ - 8y³ - 6xy(x - 2y) = 216

⇒ x³ - 8y³ - 6xy(6) = 216      {since , x - 2y = 6 from i}

⇒x³ - 8y³ - 36xy - 216 = 0

so the value of x³ - 8y³ - 36xy - 216 = 0  ANSWER

Answer 2

Given,

The polynomial x³-8y³-36xy-216 is given along with x = 2y+6.

To find,

We have to find the value of x³-8y³-36xy-216 at x = 2y+6.

Solution,

The value of x³-8y³-36xy-216 at x = 2y+6 is 0.

It is given that,

x = 2y+6

x-2y = 6        (1)

Cubing both sides of equation (1)

(x-2y)³ = 216

x³ - 8y³ - 6xy(x-2y) = 216

Using equation (1), we get

x³ - 8y³ -6xy (6) = 216

x³ - 8y³ -36 xy = 216

x³ - 8y³ -36 xy -216 = 0

Hence, the value of x³-8y³-36xy-216 at x = 2y+6 is 0.