(vi)(− 6 )^2(
−3
2) 3
=(− 1 )^262 (− 1 )^333
23=−6233
23
on using(− 1 )^2 =1,(− 1 )^3 =− 1=−( 2 × 3 )^233
23=−223233
23=−35
2
(vii) If it helps, just think ofaandbas given numbers:(−ab^2 )^3
a^2 b=(− 1 )^3 a^3 b^6
a^2 b
=−ab^5(viii) 2^2( 1
2)− 3
= 22 ( 2 −^1 )−^3= 2223 = 25
The steps to watch out for in such problems are the handling of the
minus signs and brackets, and dealing with the negative powers and recip-
rocals.
Don’t forget thatan,n≥0, is not defined fora=0.
B. We can get a long way simply by using
√
ab=√
a√
b(i)√
50 =√
25 × 2 =√
52√
2 = 5√
2
(ii)√
72 −√
8 =√
36 × 2 −√
4 × 2 = 6√
2 − 2√
2 = 4√
2
(iii)(√
27 )^3 =( 3√
3 )^3 = 33 (√
3 )^3 = 333√
3 = 34√
3 = 81√
3(iv)(√
2√
3
4) 2
=(√
3
2√
2) 2
=3
4 × 2=3
8(v)√
3√
7
√
84=√
21
√
4 × 21=√
21
2√
21=1
2(vi) To simplify√
3 + 2√
2
√
3 −√
2werationaliseit by removing all surds
from the denominator. To do this we use the algebraic identity:(a−b)(a+b)≡a^2 −b^2(see Section 2.2.1) and the removal of surds by squaring. For a√
3 −√
2 on the bottom we multiply top and bottom by√
3 +√
2,
using:(√
3 −√
2 )(√
3 +√
2 )=(√
3 )^2 −(√
2 )^2 = 3 − 2 = 1