What is expand ((3x+1)(3x-1))/9 ?

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

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expand ((3x+1)(3x-1))/9

expand

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

x^{2} - \frac{1}{9}

Solution steps

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

Expand

(3 x + 1) (3 x - 1) : 9 x^{2} - 1

= \frac{9 x ^{2} - 1}{9}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{9 x ^{2} - 1}{9} = \frac{9 x ^{2}}{9} - \frac{1}{9}

= \frac{9 x ^{2}}{9} - \frac{1}{9}

Divide the numbers:

\frac{9}{9} = 1

= x^{2} - \frac{1}{9}

2.12 + 6

(3 x - 4 y = 4, 1) (11 x + 6 y = 47)

- 2 x + 4 + 5 x - 8

\frac{5}{s ( 2 s ^{2} + 2 )}

4 b^{2} + 9 b - 3 b

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