What is expand ((1-x)/3)^2 ?

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

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expand ((1-x)/3)^2

expand

(\frac{1 - x}{3})^{2}

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

Solution steps

(\frac{1 - x}{3})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 1 - x ) ^{2}}{3 ^{2}}

(1 - x)^{2} = 1 - 2 x + x^{2}

= \frac{1 - 2 x + x ^{2}}{3 ^{2}}

3^{2} = 9

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

Apply the fraction rule:

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

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

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

Rewrite in standard form

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

(1 - \frac{1}{e ^{0.4 x}}) d x

\frac{( 2 x + 1 )}{( x ^{2} + x - 4 ) ^{3}}

(1 + 2 e^{- 2 t})^{2}

\frac{1}{3 ( x + 5 )}

\frac{2 ( n + 1 )}{n}

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