Core Connections Integrated III, 2015
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Core Connections Integrated III, 2015 View details
1. Section 9.1
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Exercise 33 Page 443

Practice makes perfect
a
To factor the given expression, we will apply the sum of cubes. a^3+b^3=(a+b)(a^2-ab+b^2) Let's start by factoring out the GCF.
x^5+8x^2y^3
x^2(x^3+8y^3)
x^2(x^3+2^3y^3)
x^2(x^3+(2y)^3)
Next, let's apply the sum of cubes formula to factor the expression further.
x^2(x^3+(2y)^3)

a^3+b^3 = (a+b)(a^2-ab+b^2)

x^2(x+2y)((x^2-x(2y)+(2y)^2)
x^2(x+2y)((x^2-x(2y)+2^2y^2)
x^2(x+2y)(x^2-2xy+4y^2)
b
To factor the given expression, we will apply the difference of two cubes. a^3-b^3 ⇔ (a-b)(a^2+ab+b^2) Let's rewrite our expression so that it matches the above form.
8y^6-125x^3
2^3y^6-3^5x^3
2^3y^(2(3))-3^5x^3
2^3(y^2)^3-3^5x^3
(2y^2)^3-(3x)^3
Not that we have written the expression, let's use the difference of cubes formula.
(2y^2)^3-(3x)^3

a^3-b^3 = (a-b)(a^2+ab+b^2)

(2y^2-3x)( (2y^2)^2+(2y^2)(3x)+(3x)^2)
(2y^2-3x)(2^2(y^2)^2+(2y^2)(3x)+3^2x^2)
(2y^2-3x)(2^2y^4+(2y^2)(3x)+3^2x^2)
(2y^2-3x)(4y^4+6xy^2+9x^2)
c
To factor the given expression, we will apply the difference of squares formula. a^2-b^2 ⇔ (a-b)(a+b)Let's rewrite our expression so that it matches the above form. x^6-y^6 ⇔ (x^3)^2-(y^3)^2 Not that we have written the expression, let's use the difference of squares formula. (x^3)^2-(y^3)^2 ⇕ (x^3-y^3)(x^3+y^3) Now, notice that the two factor are the difference and the sum of two cubes. Therefore, we can use the formulas we have used in Part A and Part B to simplify this expression further.
(x^3-y^3)(x^3+y^3)

a^3-b^3 = (a-b)(a^2+ab+b^2)

(x-y)(x^2+xy+y^2)(x^3+y^3)

a^3+b^3 = (a+b)(a^2-ab+b^2)

(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)