### Video Transcript

Complete the following 14π₯ cubed π¦ to the power of five over something is equal to two π₯ over six π¦.

When simplifying fractions, we need to remember the following rule. Whatever you do to the top, you must do to the bottom. Whatever value we divide the numerator by, we must divide the denominator by the same value. Letβs firstly consider what we need to do to get from 14π₯ cubed π¦ to the power of five to two π₯. The easiest way to do this is to divide 14π₯ cubed π¦ to the power of five by two π₯.

In order to simplify this expression, we need to remember one of our laws of indices, or exponents. π₯ to the power of π divided by π₯ to the power of π is equal to π₯ to the power of π minus π. Two π₯ is the same as two π₯ to the power of one. If we consider the numbers, we have 14 on the top and two on the bottom. 14 divided by two is equal to seven. Using the law of exponents mentioned, π₯ cubed divided by π₯ to the power of one is equal to π₯ squared. We are also left with π¦ to the power of five, as there is no π¦ term on the denominator.

14π₯ cubed π¦ to the power five divided by two π₯ is equal to seven π₯ squared π¦ to the power of five. We can therefore say that to get from the first numerator, 14π₯ cubed π¦ to the power of five, to the second numerator, two π₯, we have divided by seven π₯ squared π¦ to the power of five. The opposite of dividing is multiplying. To work out the missing term, we need to multiply six π¦ by seven π₯ squared π¦ to the power of five.

Another one of our laws of exponents, or indices, says that π₯ to the power of π multiplied by π₯ to the power of π is equal to π₯ to the power of π plus π. We can use this law to help us multiply six π¦ by seven π₯ squared π¦ to the power of five. Six multiplied by seven is equal to 42. There is only one π₯ term; this is π₯ squared. Finally, we need to multiply π¦, or π¦ to the power of one, by π¦ to the power of five. Adding the exponents gives us π¦ to the power of six. The missing term is 42π₯ squared π¦ to the power of six. This is because 14π₯ cubed π¦ to the power of five over 42π₯ squared π¦ to the power of six simplifies to two π₯ over six π¦.