Question

1f x = 2y + 6, prove that the value of x^3-8y^3 - 36xy - 212 = 4.​

Answer 1

Answer:

hola mate

See the attachment

hope it helps u

Answer 2

Answer:

Step-by-step explanation:

Given ,

x= 2y+6

To prove ,

$x^{3}-8y^{3}-36xy-216=0$

On taking cube terms ,we have

$x^{3}=(2y+6)^{3} => (a+b)^{3} =a^{3}+b^{3}+3ab(a+b)$

$x^{3} = 8y^{3}+216+36y(2y+6)$

$x^{3}=8y^{3}-216+36xy=0$

Hope it helps :)