What is ((9x^3-9x))/((-18x^3+18x)) ?

The solution to ((9x^3-9x))/((-18x^3+18x)) is -1/2

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simplify ((9x^3-9x))/((-18x^3+18x))

simplify

\frac{( 9 x ^{3} - 9 x )}{( - 18 x ^{3} + 18 x )}

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

\frac{9 x ^{3} - 9 x}{- 18 x ^{3} + 18 x}

Factor

9 x^{3} - 9 x : 9 x (x + 1) (x - 1)

= \frac{9 x ( x + 1 ) ( x - 1 )}{- 18 x ^{3} + 18 x}

Factor

- 18 x^{3} + 18 x : - 18 x (x + 1) (x - 1)

= \frac{9 x ( x + 1 ) ( x - 1 )}{- 18 x ( x + 1 ) ( x - 1 )}

Cancel the common factor:

x

= \frac{9 ( x + 1 ) ( x - 1 )}{- 18 ( x + 1 ) ( x - 1 )}

Cancel the common factor:

x + 1

= \frac{9 ( x - 1 )}{- 18 ( x - 1 )}

Cancel the common factor:

x - 1

= \frac{9}{- 18}

Cancel the numbers:

\frac{9}{- 18} = \frac{1}{- 2}

= \frac{1}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{2}

lo g_{10} (a^{2}) b^{3} c^{5}

\frac{( b ^{\frac{1}{3}} )}{( b ^{- \frac{3}{2}} \cdot b ^{\frac{1}{2}} )}

\frac{4}{3} \cdot n^{5} - \frac{1}{3} \cdot n^{4} + \frac{2}{3} \cdot n^{3} - \frac{5}{3} \cdot n^{2}

(2 x + 5)^{3} + (5 x - 3)^{3}

2 \frac{\frac{x}{x}}{50} + \frac{x}{40}

What is ((9x^3-9x))/((-18x^3+18x)) ?
The solution to ((9x^3-9x))/((-18x^3+18x)) is -1/2