What is expand (2-1/3 x)^4 ?

The solution to expand (2-1/3 x)^4 is 16-(32x)/3+(8x^2)/3-(8x^3)/(27)+(x^4)/(81)

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expand (2-1/3 x)^4

expand

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

16 - \frac{32 x}{3} + \frac{8 x ^{2}}{3} - \frac{8 x ^{3}}{27} + \frac{x ^{4}}{81}

Solution steps

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

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2, b = - \frac{1}{3} x

= i = 0 \sum 4 (i 4) \cdot 2^{(4 - i)} (- \frac{1}{3} x)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 2^{4} (- \frac{1}{3} x)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 2^{3} (- \frac{1}{3} x)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 2^{2} (- \frac{1}{3} x)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 2^{1} (- \frac{1}{3} x)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 2^{0} (- \frac{1}{3} x)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 2^{4} (- \frac{1}{3} x)^{0} : 16

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 2^{3} (- \frac{1}{3} x)^{1} : - \frac{32 x}{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 2^{2} (- \frac{1}{3} x)^{2} : \frac{8 x ^{2}}{3}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 2^{1} (- \frac{1}{3} x)^{3} : - \frac{8 x ^{3}}{27}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 2^{0} (- \frac{1}{3} x)^{4} : \frac{x ^{4}}{81}

= 16 - \frac{32 x}{3} + \frac{8 x ^{2}}{3} - \frac{8 x ^{3}}{27} + \frac{x ^{4}}{81}

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What is expand (2-1/3 x)^4 ?
The solution to expand (2-1/3 x)^4 is 16-(32x)/3+(8x^2)/3-(8x^3)/(27)+(x^4)/(81)