Question

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

Answer is "0"

Please help​

Answer 1

Answer:

$= {x}^{3} - 8 {y}^{3} - 36xy - 216 \\ = ({2y + 6})^{3} - 8 {y}^{3} - 36(2y + 6)y - 216 \\ = (8 {y}^{3} + 3 \times 2y \times 6(2y + 6) + 216) - 8 {y}^{3} - 72 {y}^{2} - 216y - 216 \\ = 36y(2y + 6) - 72 {y}^{2} - 216y \\ = 72 {y}^{2} + 216y - 72 {y}^{2} - 216y \\ = 0$heres your answer

Answer 2

Step-by-step explanation:

$\huge\underline\purple{\mathfrak</p><p>Your Answer}$

According to your question,we have:-

x³-8y³-36xy-216

=>> x³+(-8y³) + (-216) -36xy

=>> x³+ (-2y)³ + (-6)³ - 3*x(-2y) * (-6)

=>> By a³+b³+c³ -3abc ,where a=X , b= (-2y) and C=(-6)

=>> (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

since it is given that [ x-2y-6=0)

Hence, x³-8y³-36xy-216= 0 {Answer}

________Proved_______

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$\huge\underline\purple{\mathfrak</strong></p><p><strong>Thanks</strong><strong>}$