if 2x-5y= 7z-3y then ,for xyz is not equal to zero , find(8 x^3 -343y^3 -8z^3)/(xyz)=
Answers
Answer:
let's start with the assumption that
y=1 and z=1
so putting them in the given equation of
2x-5y= 7z-3y2x−5y=7z−3y
we get
2x-5(1)= 7(1)-3(1)2x−5(1)=7(1)−3(1)
from this we get
2x-5= 7-32x−5=7−3
2x-5= 42x−5=4
2x= 4+52x=4+5
2x= 92x=9
dividing by 2 we get the value of x as
x=4.5x=4.5
so we got the values of x, y, and z as
\begin{gathered}x=4.5\\ y=1\\z=1\end{gathered}
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}
xyz
x
3
−y
3
−z
3
=
4.5×1×1
(4.5)
3
−1
3
−1
3
solving the right hand side we get
\frac{x^3-y^3-z^3}{xyz}=\frac{91.125-1-1}{4.5}
xyz
x
3
−y
3
−z
3
=
4.5
91.125−1−1
\frac{x^3-y^3-z^3}{xyz}=\frac{91.125-2}{4.5}
xyz
x
3
−y
3
−z
3
=
4.5
91.125−2
\frac{x^3-y^3-z^3}{xyz}=\frac{89.125}{4.5}
xyz
x
3
−y
3
−z
3
=
4.5
89.125
hence we get the answer as
\bf{\frac{x^3-y^3-z^3}{xyz}=19.8}
xyz
x
3
−y
3
−z
3
=19.8
This of course a non trivial solution as we had three variables and one equation.