What is (c^2+6c+9)/(c^3+27)*(c^2-3c+9)/(3c+9) ?

The solution to (c^2+6c+9)/(c^3+27)*(c^2-3c+9)/(3c+9) is 1/3

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simplify (c^2+6c+9)/(c^3+27)*(c^2-3c+9)/(3c+9)

simplify

\frac{c ^{2} + 6 c + 9}{c ^{3} + 27} \cdot \frac{c ^{2} - 3 c + 9}{3 c + 9}

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

\frac{c ^{2} + 6 c + 9}{c ^{3} + 27} \cdot \frac{c ^{2} - 3 c + 9}{3 c + 9}

Cancel

\frac{c ^{2} + 6 c + 9}{c ^{3} + 27} : \frac{c + 3}{c ^{2} - 3 c + 9}

= \frac{c + 3}{c ^{2} - 3 c + 9} \cdot \frac{c ^{2} - 3 c + 9}{3 c + 9}

Cross-cancel common factor:

c^{2} - 3 c + 9

= \frac{c + 3}{3 c + 9}

Factor

3 c + 9 : 3 (c + 3)

= \frac{c + 3}{3 ( c + 3 )}

Cancel the common factor:

c + 3

= \frac{1}{3}

x^{4} + 2 x^{3} - 4 x^{2} - 8 x \geq 0

∣ - x + 2 ∣ - 13 > 17

\frac{1 - a}{a ^{2} - a + 1} + \frac{a ^{2}}{a ^{3} + 1}

\frac{31}{\frac{25}{\frac{12}{25}}}

w + w + (2 w - 2) + (2 w - 2) = 11

What is (c^2+6c+9)/(c^3+27)*(c^2-3c+9)/(3c+9) ?
The solution to (c^2+6c+9)/(c^3+27)*(c^2-3c+9)/(3c+9) is 1/3