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Find the product of $(-3 x y z)(\frac{4}{9} x^{2} z)(-\frac{27}{2} x y^{2} z)$ and verify the result for ; $x=2, y=3$ and $z=-1$
Given :
The given expression is $(-3 x y z)(\frac{4}{9} x^{2} z)(-\frac{27}{2} x y^{2} z)$.
To do :
We have to find the product of the given expression and verify the result for $x=2, y=3$ and $z=-1$.
Solution :
$(-3 x y z)(\frac{4}{9} x^{2} z)(-\frac{27}{2} x y^{2} z)$
$(-3 x y z)(\frac{4}{9} x^{2} z)(-\frac{27}{2} x y^{2} z) = (-3 \times \frac{4}{9}\times \frac{-27}{2}) \times (x \times x^2 \times x) \times (y \times y^2) \times (z \times z \times z)$
$ = (-3 \times 2 \times -3) \times x^{1+2+1} \times y^{1+2} \times z^{1+1+1}$
$ = 18x^4 y^3 z^3$
If $x = 2, y = 3$ and $z = -1$ then $(-3 x y z)(\frac{4}{9} x^{2} z)(-\frac{27}{2} x y^{2} z) = $
$ = 18x^4 y^3 z^3 = 18 (2)^4 (3)^3 (-1)^3$
$ = 18(16)(9)(-1)$
$= -2592$.
Therefore, the value of $(-3 x y z)(\frac{4}{9} x^{2} z)(-\frac{27}{2} x y^{2} z)$ at $x = 2, y = 3$ and $z = -1$ is $-2592$
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