(ab)n=anbn

a^0 = 1

wherea,bare algebraic functions,m,nneed not be integers,butcare is needed when
they are not. For example

√
√ x^2 +1 exists for all values ofx,sincex^2 +1 is positive, but
x^2 −1 only exists (as a real number) ifx^2 ≥1, i.e.x≥1orx≤−1.
As always we need to be careful when brackets and signs are involved, for example
(− 2 x)^3 =(− 1 )^323 x^3 =− 8 x^3.

Solution to review question 2.1.12

(i)

a^2 c^4

a−^3 b^2 c

=a^2 c^4 (a−^3 )−^1 b−^2 c−^1 =a^2 a^3 b−^2 c^4 c−^1

=a^5 b−^2 c^3

(ii)

( 3 x)^3 ( 2 y)−^1

(x^2 y)−^1

= 33 x^32 −^1 y−^1 x^2 y

=

27

2

x^5

(iii) Here, treat the algebraic expressions under the indices as single objects.

(x− 1 )

1

(^4) ( 16 x+ 16 )
1
4
(x+ 1 )(x^2 − 1 )−
3
4

(x− 1 )
1
(^4) ( 16 (x+ 1 ))
1
(^4) (x^2 − 1 )
3
4
(x+ 1 )

16
1
(^4) (x− 1 )
1
(^4) (x+ 1 )
1
(^4) (x^2 − 1 )
3
4
x+ 1

16
1
(^4) (x^2 − 1 )
1
(^4) (x^2 − 1 )
3
4
x+ 1

16
1
(^4) (x^2 − 1 )
x+ 1
= 2 (x− 1 )
Note that in these formal manipulations using the rules of indices we have
to remember that for real quantities throughout we must havex>1.
2.2.13 Binomial theorem
➤
40 79 ➤
There are many occasions in applied mathematics when we need to expand expressions
such as(a+b)^6. Very simple examples can be done long hand:
(a+b)^2 =a^2 + 2 ab+b^2
(a+b)^3 =(a^2 + 2 ab+b^2 )(a+b)
=a^3 + 3 a^2 b+ 3 ab^2 +b^3