Question

2 X + 3 Y + Z = 0 then 8 x cube + 27 y cube + Z Cube ÷xyz​

Answer 1

Question:

 if   2 X + 3 Y + Z = 0 then  find  8 x cube + 27 y cube + Z Cube ÷xyz​

Given:

$2x+3y+z=0$

To find:

 $\dfrac{8x^3+27y^3+z^3}{xyz}$

Solving Question:

  For answer,here we have to recall an identity.

x³+y³+z³-3xyz= (x+y+z)(x²+y²+z²-xy-yz-xz)

Solution:

let's substitute the given equation in the identity

⇒ 2x³+27y³+z³- 3xyz = ( 2x +3y +z)(4x² +9y²+z² .......)

⇒  2x³+27y³+z³-3xyz = ( 0 ) (4x² +9y²+z² .......)

⇒ 2x³+27y³+z³-3xyz = 0

⇒ 2x³+27y³+z³ = 3xyz ...........equ(1)

$\dfrac{8x^3+27y^3+z^3}{xyz}\\\\substitute\:the\:value\:equ(1)\\\\\implies \dfrac{3xyz}{xyz}\\\\\implies3*\dfrac{xyz}{xyz}\\\\\implies3*1=3$

Hence, the value of$\dfrac{8x^3+27y^3+z^3}{xyz}$is 3

Answer 2

Answer:

Step-by-step explanation:

2x+3y+z=0----Given

2x+3y=-z--------(1)

Cubing both sides we get,

8x^3+27y^3+18xy(2x+3y)=-z^3 (since,(a+b)^3=a^3+b^3+3ab(a+b))

8x^3+27y^3+z^3 =-18xy(2x+3y)

From equation 1-

2x+3y=-z

therefore,

8x^3+27y^3+z^3=-18xy*(-z)

8x^3+27y^3+z^3=18xyz

dividing both sides by xyz,we get

(8x^3+27y^3+z^3)/xyz=18

Hence our answer is 18 and i quite sure about it.

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