Question

1. which are valid factorizations of ( 729x^3 - 125 )?

a. ( left(9x + \frac{5}{6}
ight)left(81x^2 - \frac{15}{3}x + \frac{125}{1296}
ight) )
b. ( left(9x + \frac{5}{6}
ight)left(81x^2 + \frac{15}{3}x - \frac{125}{36}
ight) )
c. ( left(9x - \frac{5}{6}
ight)left(81x^2 + \frac{15}{3}x + \frac{125}{1296}
ight) )
d. ( left(9x - \frac{5}{6}
ight)left(81x^2 + \frac{15}{2}x + \frac{25}{36}
ight) )

Explanation:

Step1: Recognize difference of cubes

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

Step2: Identify $a$ and $b$

$729x^3=(9x)^3$, so $a=9x$; $\frac{125}{216}=(\frac{5}{6})^3$, so $b=\frac{5}{6}$

Step3: Compute $a^2$, $ab$, $b^2$

$a^2=(9x)^2=81x^2$, $ab=9x\cdot\frac{5}{6}=\frac{15}{2}x$, $b^2=(\frac{5}{6})^2=\frac{25}{36}$

Step4: Substitute into formula

$(9x - \frac{5}{6})(81x^{2} + \frac{15}{2}x + \frac{25}{36})$

Answer:

c. $(9x - \frac{5}{6})(81x^{2} + \frac{15}{2}x + \frac{25}{36})$