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

The solution to expand 1/3 (1-x+(xx)/2)^3 is (x^6}{24}-\frac{x^5)/4+(3x^4)/4-(4x^3)/3+(3x^2)/2-x+1/3

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

expand

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

\frac{x ^{6}}{24} - \frac{x ^{5}}{4} + \frac{3 x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2} - x + \frac{1}{3}

Solution steps

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

(1 - x + \frac{xx}{2})^{3} = \frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{8} + 1

= \frac{1}{3} (\frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{8} + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = \frac{1}{3}, b = \frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{8}, c = 1

= \frac{1}{3} \cdot \frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{8} + \frac{1}{3} \cdot 1

= \frac{1}{3} \cdot \frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{8} + 1 \cdot \frac{1}{3}

Simplify

\frac{1}{3} \cdot \frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{8} + 1 \cdot \frac{1}{3} : \frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{24} + \frac{1}{3}

= \frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{24} + \frac{1}{3}

Apply the fraction rule:

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

\frac{x ^{6} - 6 x ^{5} + 18 x ^{4} - 32 x ^{3} + 36 x ^{2} - 24 x}{24} = \frac{x ^{6}}{24} - \frac{6 x ^{5}}{24} + \frac{18 x ^{4}}{24} - \frac{32 x ^{3}}{24} + \frac{36 x ^{2}}{24} - \frac{24 x}{24}

= \frac{x ^{6}}{24} - \frac{6 x ^{5}}{24} + \frac{18 x ^{4}}{24} - \frac{32 x ^{3}}{24} + \frac{36 x ^{2}}{24} - \frac{24 x}{24} + \frac{1}{3}

\frac{6 x ^{5}}{24} = \frac{x ^{5}}{4}

\frac{18 x ^{4}}{24} = \frac{3 x ^{4}}{4}

\frac{32 x ^{3}}{24} = \frac{4 x ^{3}}{3}

\frac{36 x ^{2}}{24} = \frac{3 x ^{2}}{2}

\frac{24 x}{24} = x

= \frac{x ^{6}}{24} - \frac{x ^{5}}{4} + \frac{3 x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2} - x + \frac{1}{3}

\frac{5 ( 3 - 13 )}{4}

\frac{( e ^{- 7 s} )}{( s - 1 ) ( s - 3 )}

(x + e^{\frac{x}{2}} y)^{2}

\frac{1}{( cos ( θ ) + i sin ( θ ) ) ^{2}}

(4 x^{3} - 3 x)^{4}

What is expand 1/3 (1-x+(xx)/2)^3 ?
The solution to expand 1/3 (1-x+(xx)/2)^3 is (x^6}{24}-\frac{x^5)/4+(3x^4)/4-(4x^3)/3+(3x^2)/2-x+1/3