Simplify$\left(\frac{1}{2}{a}^2{b}^2\right){}^3\left[{\left(-4b\right)}^2\right]^2$

If a term with a power is itself raised to a power, then the powers are multiplied together.

In general:${\left(x^m\right)}^n=x^{m\times n}$

If the two terms having the same base are multiplied together, then their indices are added.

In general:$x^m\times x^n=x^{m+n}$

According to this lawwhen a product is raised to a power, every factor of the product is raised to the power.

In general:${\left(xy\right)}^m=x^m{y}^m$

If a fraction is raised to a power, then every terms of the fraction is raised to the power.

In general:${\left(\frac{x}{y}\right)}^m=\frac{x^m}{y^m}$

The given expression is:$\left(\frac{1}{2}{a}^2{b}^2\right){}^3\left[{\left(-4b\right)}^2\right]^2$

Apply the laws of indices and simplify the above expression.

$$\begin{gathered} \left(\frac{1}{2}{a}^2{b}^2\right){}^3\left[{\left(-4b\right)}^2\right]^2={\left(\frac{1}{2}\right)}^3{\left(a^2\right)}^3{\left(b^2\right)}^3{\left[{\left(-4b\right)}^2\right]}^2 \\ =\left(\frac{1^3}{2^3}\right)\left(a^{2\times 3}\right)\left(b^{2\times 3}\right)\left[{\left(-4b\right)}^{2\times 2}\right] \\ =\left(\frac{1}{8}\right)\left(a^6\right)\left(b^6\right)\left[{\left(-4b\right)}^4\right] \\ =\left(\frac{1}{8}{a}^6{b}^6\right)\left[{\left(-4\right)}^4\left(b^4\right)\right] \\ =\left(\frac{1}{8}{a}^6{b}^6\right)\left(256b^4\right) \end{gathered}$$

Collect the like terms together and simplify further.

$$\begin{gathered} \left(\frac{1}{2}{a}^2{b}^2\right){}^3\left[{\left(-4b\right)}^2\right]^2=\left(\frac{1}{8}\times 256\right)\left(a^6\right)\left(b^6\times b^4\right) \\ =\left(32\right)\left(a^6\right)\left(b^{6+4}\right) \\ =32a^6{b}^{10} \end{gathered}$$