Question
Simplify, giving Solution with positive index
$(4x^2y^3)^3 ÷ (3x^2y^3)^3$
$(4x^2y^3)^3 ÷ (3x^2y^3)^3$
✓
Answer
$\left(4 x^2 y^3\right)^3 \div\left(3 x^2 y^3\right)^3 $
$ =\frac{4^3 x^{2 \times 3} y^{3 \times 3}}{3^3 x^{2 \times 3} y^{3 \times 3}} $
$ =\frac{4^3 x^6 y^9}{3^3 x^6 y^9} $
$=\frac{4^3}{3^3}=\frac{64}{27}$
$ =\frac{4^3 x^{2 \times 3} y^{3 \times 3}}{3^3 x^{2 \times 3} y^{3 \times 3}} $
$ =\frac{4^3 x^6 y^9}{3^3 x^6 y^9} $
$=\frac{4^3}{3^3}=\frac{64}{27}$
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