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Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248

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Section
BuyFindarrow_forward

Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248
Textbook Problem

Rewrite each expression in Exercises 1–16 as a single rational expression, simplified as much as possible.

1 ( x + y ) 3 1 x 3 y

To determine

To calculate: The simplified form of the expression 1(x+y)31x3y.

Explanation

Given Information:

The provided expression is 1(x+y)31x3y.

Formula used:

The difference of two rational expression:

ABCB=ACB

To subtract when the denominators are the same, subtract the second numerators from the first and keep the common denominators.

The algebraic identity for differences of cube of two numbers:

(a3b3)=(ab)(a2+b2+ab)

Algebraic identity for squares of sum of two numbers:

(a+b)2=a2+b2+2ab

Calculation:

Consider the expression 1(x+y)31x3y.

The denominators of the terms 1(x+y)31x3 are (x+y)3 and x3.

Therefore, multiply by x3x3 in the first term and (x+y)3(x+y)3 in the second term.

1(x+y)31x3=x3x31(x+y)3(x+y)3(x+y)31x3=x3x3(x+y)3(x+y)3x3(x+y)3

Apply the difference of two rational expression

1(x+y)31x3=x3x31(x+y)3(x+y)3(x+y)31x3=x3x3(x+y)3(x+y)3x3(x+y)3=x3(x+y)3x3(x+y)3

Now, apply the algebraic identity for difference of cube of two numbers,

1(x+y)31x3=x3x31(x+y)3(x+y)3(x+y)31x3=x3x3(x+y)3<

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