How do you simplify $\frac{{(u^{0} v^{4})}^{3} \cdot - u^{3} v^{- 2}}{- 2 u}$ $\frac { ( u ^ { 0} v ^ { 4} ) ^ { 3} \cdot - u ^ { 3} v ^ { - 2} } { - 2u }$ ?

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Answer 1 · 2017-05-26T23:52:10

$\frac{1}{2} v^{10} u^{2}$ $1/2v^10u^2$

$\frac{{(u^{0} v^{4})}^{3} \cdot - u^{3} v^{- 2}}{- 2 u}$ $((u^0v^4)^3*-u^3v^-2)/(-2u)$

$= \frac{{(1 \cdot v^{4})}^{3} \cdot u^{3} v^{- 2}}{2 u}$ $=((1*v^4)^3*u^3v^-2)/(2u)$

$= \frac{v^{12} \cdot u^{3}}{2 u \cdot v^{2}}$ $=(v^12*u^3)/(2u*v^2)$

$= \frac{1}{2} v^{10} u^{2}$ $=1/2v^10u^2$

How do you simplify (u0v4)3⋅−u3v−2−2u\frac { ( u ^ { 0} v ^ { 4} ) ^ { 3} \cdot - u ^ { 3} v ^ { - 2} } { - 2u }?