Express as a single fraction in its simplest form $\dfrac{18b^7}{5c^2} - \dfrac{3b^4}{81}$. Use the laws of indices to show that $6^4 \times 100 + 116 \times 36^2$ can be expressed as a single power of six.

1. Simplify the numerical denominator in the second fraction and choose a common denominator.

$\frac{18b^7}{5c^2}-\frac{3b^4}{81}=\frac{18b^7}{5c^2}-\frac{b^4}{27}$

The common denominator of $5c^2$ and $27$ is $135c^2$.

2. Express both terms over the common denominator and combine into one fraction.

$\frac{18b^7}{5c^2}-\frac{b^4}{27}$
$=\frac{18b^7\times 27}{135c^2}-\frac{b^4\times 5c^2}{135c^2}$
$=\frac{486b^7-5b^4c^2}{135c^2}$
$=\frac{b^4(486b^3-5c^2)}{135c^2}$

1. Express $36^2$ as a power of $6$.

Since $36 = 6^2$,
$36^2 = (6^2)^2 = 6^4.$

2. Factorise and combine using laws of indices to obtain a single power of $6$.

$6^4 \times 100 + 116 \times 36^2$
$= 100 \times 6^4 + 116 \times 6^4$
$= (100+116)6^4$
$=216 \times 6^4$
$=6^3 \times 6^4$
$=6^7.$