What is expand (k-1/3)^3 ?

The solution to expand (k-1/3)^3 is (27k^3-27k^2+9k-1)/(27)

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expand (k-1/3)^3

expand

(k - \frac{1}{3})^{3}

\frac{27 k ^{3} - 27 k ^{2} + 9 k - 1}{27}

Solution steps

(k - \frac{1}{3})^{3}

Apply Perfect Cube Formula:

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

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

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

Simplify

k^{3} - 3 k^{2} \frac{1}{3} + 3 k (\frac{1}{3})^{2} - (\frac{1}{3})^{3} : \frac{27 k ^{3} - 27 k ^{2} + 9 k - 1}{27}

= \frac{27 k ^{3} - 27 k ^{2} + 9 k - 1}{27}

A (- 5, 2) B (3, - 1)

\frac{3 ( 27 - 2 x ^{3} )}{( 27 + x ^{3} ) ^{2}}

x (2 x + 1)

(2 x^{3} + 2)^{4} \cdot cos^{3} (x)

(4.3 x^{105}) l e f t (2, 1 x^{103}) l e f t (6 x^{10} - 4)

What is expand (k-1/3)^3 ?
The solution to expand (k-1/3)^3 is (27k^3-27k^2+9k-1)/(27)