Question

which are valid factorizations of ( 729x^3 - \frac{125}{216} )?

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{225}{36}
ight) )

factor: ( 1.331x^3 - 64 ).

a. ( (1.1x + 4)(1.21x^2 - 4.4x + 16) )
b. ( (1.1x - 4)(1.21x^2 + 4.4x + 16) )
c. ( (1.1x - 4)(1.21x^2 - 4.4x - 16) )
d. ( (1.1x + 4)(1.21x^2 + 4.4x - 16) )

Explanation:

First Question:

Step1: Recognize difference of cubes

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

Step2: Rewrite terms as cubes

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

Step3: Apply difference of cubes

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

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Second Question:

Step1: Recognize difference of cubes

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

Step2: Rewrite terms as cubes

$1.331x^3=(1.1x)^3$, $64=4^3$

Step3: Apply difference of cubes

$a=1.1x$, $b=4$
$a^2=(1.1x)^2=1.21x^2$, $ab=1.1x\cdot4=4.4x$, $b^2=4^2=16$
Result: $(1.1x - 4)(1.21x^2 + 4.4x + 16)$

Answer:

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