Question

evaluate \\(dfrac{(4^{2/3})(2^{1/6})(3^{3/2})}{(2^{-1/2})(3^{1/2})}\\)

Explanation:

Step1: Rewrite 4 as $2^2$

$\frac{( (2^2)^{2/3} )(2^{1/6})(3^{3/2})}{(2^{-1/2})(3^{1/2})}$

Step2: Simplify $(2^2)^{2/3}$

$(2^2)^{2/3}=2^{4/3}$, so:
$\frac{(2^{4/3})(2^{1/6})(3^{3/2})}{(2^{-1/2})(3^{1/2})}$

Step3: Combine 2's exponents (numerator)

Add exponents: $\frac{4}{3}+\frac{1}{6}=\frac{8}{6}+\frac{1}{6}=\frac{9}{6}=\frac{3}{2}$
$\frac{2^{3/2} \cdot 3^{3/2}}{2^{-1/2} \cdot 3^{1/2}}$

Step4: Simplify 2's exponents (division)

Subtract exponents: $\frac{3}{2}-(-\frac{1}{2})=\frac{3}{2}+\frac{1}{2}=2$
$\frac{2^{2} \cdot 3^{3/2}}{3^{1/2}}$

Step5: Simplify 3's exponents (division)

Subtract exponents: $\frac{3}{2}-\frac{1}{2}=1$
$2^{2} \cdot 3^{1}$

Step6: Calculate final value

$2^2=4$, $3^1=3$, so $4 \times 3 = 12$

Answer:

12