Solve (27x^3y^3z^2)div(1/3x^2yz)*(3yz)

$$( 27 x ^ { 3 } y ^ { 3 } z ^ { 2 } ) \div ( \frac { 1 } { 3 } x ^ { 2 } y z ) \cdot ( 3 y z )$$

Evaluate

$$243xz^{2}y^{3}$$

Solution Steps

Cancel out yzx^{2} in both numerator and denominator.

$$\frac{27xzy^{2}}{\frac{1}{3}}\times 3yz$$

Divide 27xzy^{2} by \frac{1}{3} by multiplying 27xzy^{2} by the reciprocal of \frac{1}{3}.

$$27xzy^{2}\times 3\times 3yz$$

Multiply 27 and 3 to get 81.

$$81xzy^{2}\times 3yz$$

Multiply 81 and 3 to get 243.

$$243xzy^{2}yz$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$243xzy^{3}z$$

Multiply z and z to get z^{2}.

$$243xz^{2}y^{3}$$

Differentiate w.r.t. x

$$243z^{2}y^{3}$$