# the given expression ( a 1 6 b − 3 x − 1 y ) 3 ( x − 2 b − 1 a 3 2 y 1 3 ) .

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.2, Problem 70E

(a)

To determine

Expert Solution

## Answer to Problem 70E

(a16b3x1y)3(x2b1a32y13)=xab10y103

### Explanation of Solution

Given information:

Given expression

(a16b3x1y)3(x2b1a32y13)

Calculation:

Consider the given expression.

(a16b3x1y)3(x2b1a32y13)

Now, use the law of exponents, which states

(am)n=amn , (ab)n=anbn and (ab)n=anbn

And combine like terms

(a16b3x1y)3(x2b1a32y13)=(a16)3(b3)3(x1)3(y)3(x2b1a32y13)=a12b9x3y3(x2b1a32y13)=x2+3y313b91a1232 [aman=am+n,aman=amn]=x1y103b10a1=xab10y103 [an=1an]

Hence,

(a16b3x1y)3(x2b1a32y13)=xab10y103

(b)

To determine

Expert Solution

## Answer to Problem 70E

(9st)32(27s3t4)23(3s24t13)1=4s32t92

### Explanation of Solution

Given information:

Given expression

(9st)32(27s3t4)23(3s24t13)1

Calculation:

Consider the given expression.

(9st)32(27s3t4)23(3s24t13)1

Now, use the law of exponents, which states

(am)n=amn , (ab)n=anbn and (ab)n=anbn

And combine like terms

(9st)32(27s3t4)23(3s24t13)1=(9)32s32t32(27)23(s3)23(t4)2331(s2)141(t13)1=27s32t329s2t234s23t13=4s32t32+3 [aman=am+n]=4s32t92

Hence,

(9st)32(27s3t4)23(3s24t13)1=4s32t92

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