Question

factor: (729x^3 - \frac{64}{729}).
a. ((9x - \frac{4}{9})(81x^2 + \frac{12}{3}x + \frac{16}{81}))
b. ((9x - \frac{4}{9})(81x^2 - \frac{12}{3}x - \frac{16}{81}))
c. ((9x + \frac{4}{9})(81x^2 - \frac{12}{3}x + \frac{16}{81}))
d. ((9x + \frac{4}{9})(81x^2 + \frac{12}{3}x - \frac{16}{81}))

factor: (64y^3 - 0.343).
a. ((4y + 0.7)(16y^2 - 2.8y + 0.49))
b. ((4y - 0.7)(16y^2 - 2.8y - 0.49))
c. ((4y - 0.7)(16y^2 + 2.8y + 0.49))
d. ((4y + 0.7)(16y^2 + 2.8y - 0.49))

Explanation:

Step1: Identify difference of cubes

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

Step2: Solve first expression

For $729x^3 - \frac{64}{729}$:
$a=9x$, since $(9x)^3=729x^3$; $b=\frac{4}{9}$, since $(\frac{4}{9})^3=\frac{64}{729}$
Substitute into formula:
$(9x - \frac{4}{9})[(9x)^2 + (9x)(\frac{4}{9}) + (\frac{4}{9})^2]=(9x - \frac{4}{9})(81x^2 + 4x + \frac{16}{81})$
(Note: $\frac{12}{3}x$ simplifies to $4x$, matching the derived term)

Step3: Solve second expression

For $64y^3 - 0.343$:
$a=4y$, since $(4y)^3=64y^3$; $b=0.7$, since $(0.7)^3=0.343$
Substitute into formula:
$(4y - 0.7)[(4y)^2 + (4y)(0.7) + (0.7)^2]=(4y - 0.7)(16y^2 + 2.8y + 0.49)$

Answer:

1. a. $(9x - \frac{4}{9})(81x^2 + \frac{12}{3}x + \frac{16}{81})$
2. c. $(4y - 0.7)(16y^2 + 2.8y + 0.49)$