This sounds like it's going to be really difficult. But if you have a good
feeling about the first three lessons, I think you'll find that it's not too bad.
Again, the lesson is in three sections so take your time and get it right!

SECTION A

First you're going to learn how to simplify something like \((a^2)^3\) .

By the way, we say \(a\) squared to the power of 3.

Let's take it apart (unsimplify it) first, as this has worked so well for us in the past!

So \((a^2)^3~=~(a^2)\times (a^2)\times (a^2)\)

Make sure you understand this last line before you go any further.

Now let's open things up even more: \((a\times a)\times (a\times a)\times (a\times a)\)

And get rid of the brackets, which are now not needed: \(a\times a\times a\times a\times a\times a~=~a^6\)

An even more impressive method

You might have realised at the start that we were going to end up with
3 lots of 2 lots of \(a\) multiplied together, which is a total of
6 \(a\)'s multiplied together, which is \(a^6\).

Since 3 lots of 2 is another way of saying 3 \(\times\) 2,
just multiply the powers (3 \(\times\) 2 = 6)!

This method or rule is written formally like this: \((x^a)^b~=~ x^{a\times b}\) or just \(x^{ab}\)

Examples

1. Simplify \((x^5)^2\)

Answer: \(x^{5\times 2}~=~x^{10}\)

2. Simplify \((y^{12})^{7}\)

Answer: \(x^{12\times 7}~=~x^{84}\)

Practise to master

SECTION A

Simplify each of these expressions using your knowledge of indices.

01) \((m^2)^3\)

\(m^6\)

02) \((n^4)^2\)

\(n^8\)

03) \((y^3)^4\)

\(y^{12}\)

04) \((x^5)^3\)

\(x^{15}\)

05) \((g^7)^2\)

\(g^{14}\)

06) \((t^6)^5\)

\(t^{30}\)

SECTION B

Now we'll look at what happens when there is a bit more going on in the
bracket.

Before we go any further, we can already see that there are going to be
3 lots of 3 lots of \(a\) multiplied together, which is 9 lots of \(a\) multiplied together, which
is \(a^9\) . Then we're going to have 3 lots of 2 lots of \(b\) , which is
6 lots of \(b\) multiplied together, which is \(b^6\) . So the
answer is \(a^9b^6\) .

Basically, we're doing exactly the same as we did in section A, except we're
doing it to each bit in the bracket separately.

We could just write \((a^3b^2)^3~=~a^{3\times 3}b^{2\times 3}~=~a^9b^6\) .

Example 2

Simplify \((2m^3)^5\)

Answer: \(2^5\times m^{3\times 5}~=~32m^{15}\)

You could say there's an invisible power of 1 on the 2, and 1 \(\times\) 5 = 5!

You could say there's an invisible power of 1 on the 3 and the \(y\) , and 1 \(\times\) 2 = 2!

Practise to master

SECTION B

Simplify these expressions using your knowledge of indices.

01) \((xy^4)^2\)

\(x^2y^8\)

02) \((a^2b^3)^4\)

\(a^8b^{12}\)

03) \((p^4q^2)^2\)

\(p^8q^4\)

04) \((c^5d^2)^3\)

\(c^{15}d^6\)

05) \((2e^3)^4\)

\(16e^{12}\)

06) \((3x^2)^3\)

\(27x^6\)

07) \((4w^4)^2\)

\(16w^8\)

08) \((3y^7)^3\)

\(27y^{21}\)

09) \((4rs^3)^3\)

\(64r^3s^9\)

10) \((5x^3y^4)^2\)

\(25x^6y^8\)

11) \((3u^4v^2)^3\)

\(27u^{12}v^6\)

12) \((2g^2h^6)^4\)

\(16g^8h^{24}\)

SECTION C

OK, I'm going to throw in one last complication and then we're done! What
if we have bits inside and bits outside the brackets?

Example

Simplify \(2x(3x^3y)^2\)

This means \(2x\times (3x^3y)^2\) and the Order of Operations
(BIDMAS, BODMAS, whatever) tells us to deal with indices before multiplication.

If we do this, we get \(2x\times 9x^6y^2\)

Answer: \(18x^7y^2\)

The important things to remember here are:

(a) Only raise terms that are IN the brackets to the power outside the brackets.

(b) Do this BEFORE you multiply by the terms outside the brackets.

A note about workings - you DO need them! They may earn you marks in an exam
but, more importantly, they will help you to avoid making mistakes.
I've also included workings with the answers below.