Question

x cube minus 27 y cube plus 8 z cube plus 18xyz​

Answer 1

Step-by-step explanation:

${x}^{3} - 27 {y}^{3} + 8 {z}^{3} + 18xyz$

we know that if a+b+c=0

then

$a ^{3} + b ^{3} + c ^{3} = 3abc$

HERE a=x

b=-3y

c=2z

$= > {x}^{3} - 27 {y}^{3} + 8 {z}^{3} = 3 \times (x) \times ( - 3y) \times (2z)$

$= > {x}^{3} - 27 {y}^{3} + 8 {z}^{3} = 3 \times (x) \times ( - 3y) \times (2z)$

$= > {x}^{3} - 27 {y}^{3} + 8 {z}^{3} = - 18xyz$

$= > {x}^{3} - 27 {y}^{3} + 8 {z}^{3} + 18xyz =0$

THEREFORE ANSWER IS ZERO

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