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

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

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expand (2-x/3)^4

expand

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

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

Solution steps

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

Apply binomial theorem:

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

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

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

7 e^{5 t + 2} + 5 e^{(3 + t)^{2}}

- \frac{4}{s ( s ^{2} + 2 s + 2 )}

lo g_{5} (x + 2)

(x + 5 i)^{2}

\frac{( 3 + y ) ^{2}}{9}

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