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

The solution to expand (z-1/3)^4 is z^4-(4z^3)/3+(2z^2)/3-(4z)/(27)+1/81

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expand (z-1/3)^4

expand

(z - \frac{1}{3})^{4}

z^{4} - \frac{4 z ^{3}}{3} + \frac{2 z ^{2}}{3} - \frac{4 z}{27} + \frac{1}{81}

Solution steps

(z - \frac{1}{3})^{4}

Apply binomial theorem:

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

a = z, b = - \frac{1}{3}

= i = 0 \sum 4 (i 4) z^{(4 - i)} (- \frac{1}{3})^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} z^{4} (- \frac{1}{3})^{0} : z^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} z^{3} (- \frac{1}{3})^{1} : - \frac{4 z ^{3}}{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} z^{2} (- \frac{1}{3})^{2} : \frac{2 z ^{2}}{3}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} z^{1} (- \frac{1}{3})^{3} : - \frac{4 z}{27}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} z^{0} (- \frac{1}{3})^{4} : \frac{1}{81}

= z^{4} - \frac{4 z ^{3}}{3} + \frac{2 z ^{2}}{3} - \frac{4 z}{27} + \frac{1}{81}

5 (6 + 7)

x^{\frac{1}{3}} (x + 8)

\frac{d}{d x} ((y) (x^{3} + a x^{2} + b x + c))

x + h \cdot (- x)

x^{3} y d x + (x^{4} + x^{2} y^{2} + y^{4}) d y

What is expand (z-1/3)^4 ?
The solution to expand (z-1/3)^4 is z^4-(4z^3)/3+(2z^2)/3-(4z)/(27)+1/81