Question

if 2x-5y= 7z-3y then ,for xyz is not equal to zero , find( x^3 -y^3 -z^3)/(xyz)=

Answer 1

let's start with the assumption that

y=1 and z=1

so putting them in the given equation of

$2x-5y= 7z-3y$

we get

$2x-5(1)= 7(1)-3(1)$

from this we get

$2x-5= 7-3$

$2x-5= 4$

$2x= 4+5$

$2x= 9$

dividing by 2 we get the value of x as

$x=4.5$

so we got the values of x, y, and z as

$x=4.5\\ y=1\\z=1$

putting them in the given question we get

$\frac{x^3-y^3-z^3}{xyz}=\frac{(4.5)^3-1^3-1^3}{4.5\times 1 \times 1}$

solving the right hand side we get

$\frac{x^3-y^3-z^3}{xyz}=\frac{91.125-1-1}{4.5}$

$\frac{x^3-y^3-z^3}{xyz}=\frac{91.125-2}{4.5}$

$\frac{x^3-y^3-z^3}{xyz}=\frac{89.125}{4.5}$

hence we get the answer as

$\bf{\frac{x^3-y^3-z^3}{xyz}=19.8}$

This of course a non trivial solution as we had three variables and one equation.

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