$$ \frac{\left(4 x^{2} y^{3}\right)^{2}}{x^{2} y} $$ as $$ x=-2 $$ en $$ y=-1 $$

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Substitute $$ x=-2,y=-1 $$ into the expression $$ \frac{(4x^2y^3)^2}{x^2y} $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$\frac{\left(4x^{2}y^{3}\right)^{2}}{x^{2}y}$$
- step1: Substitute:
  $$\frac{\left(4\left(-2\right)^{2}\left(-1\right)^{3}\right)^{2}}{\left(-2\right)^{2}\left(-1\right)}$$
- step2: Multiply the terms:
  $$\frac{\left(-16\right)^{2}}{\left(-2\right)^{2}\left(-1\right)}$$
- step3: Multiply the terms:
  $$\frac{\left(-16\right)^{2}}{-2^{2}}$$
- step4: Rewrite the expression:
  $$\frac{16^{2}}{-2^{2}}$$
- step5: Rewrite the expression:
  $$\frac{2^{8}}{-2^{2}}$$
- step6: Reduce the fraction:
  $$\frac{2^{6}}{-1}$$
- step7: Calculate:
  $$-2^{6}$$
- step8: Evaluate the power:
  $$-64$$
The value of $$ \frac{\left(4 x^{2} y^{3}\right)^{2}}{x^{2} y} $$ when $$ x=-2 $$ and $$ y=-1 $$ is -64.
When $$ x=-2 $$ and $$ y=-1 $$, the expression $$ \frac{\left(4 x^{2} y^{3}\right)^{2}}{x^{2} y} $$ equals -64.

\( \frac{\left(4 x^{2} y^{3}\right)^{2}}{x^{2} y} \) as \( x=-2 \) en \( y=-1 \)

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