Question

solve:x³-8y³-36xy-216 if x=2y+6

plz answers if u are sure..​

Answer 1

Step-by-step explanation:

x³-8y³-36xy-216

=(2y+6)³-8y³-36(2y+6)y-216

=(2y) ³+6³+3×(2y)²×6+3×2y×6²-8y³-36(2y²+6y) -216

=8y³+216+72y²+216y-8y³-72y²-216y-216

=0

Answer 2

Step-by-step explanation:

Given that if x=2y+6

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

we can write this as

${x}^{3} + {( - 2y)}^{3} + {( - 6)}^{3} - 3( - 2y)( - 6)x$

Now, this is equal to

$(x - 6 - 2y)( {x}^{2} + 4 {y}^{2} + 36 + 2xy - 12y + 6x)$

now, (2y+6-6-2y) =0

hence

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

I hope this will help you..... thanks