Question

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

Answer 1

Put x = 2y + 6 in x3-8y3-36xy-216
∴ The given polynomial becomes, (2y+6)3�– 8y3�– 36(2y+6)y – 216
= 8y3�+ 72y2�+ 216y + 216 – 8y3�– 72y2�– 216y – 216�� [Since (a+b)3�= a3�+ 3a2b + 3ab2�+ b3]
= 0

Answer 2

Answer:

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