What is expand (((2-y)^3))/3 ?

The solution to expand (((2-y)^3))/3 is -(y^3)/3+2y^2-4y+8/3

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expand (((2-y)^3))/3

expand

\frac{( ( 2 - y ) ^{3} )}{3}

- \frac{y ^{3}}{3} + 2 y^{2} - 4 y + \frac{8}{3}

Solution steps

\frac{( 2 - y ) ^{3}}{3}

(2 - y)^{3} = 8 - 12 y + 6 y^{2} - y^{3}

= \frac{8 - 12 y + 6 y ^{2} - y ^{3}}{3}

Apply the fraction rule:

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

\frac{8 - 12 y + 6 y ^{2} - y ^{3}}{3} = \frac{8}{3} - \frac{12 y}{3} + \frac{6 y ^{2}}{3} - \frac{y ^{3}}{3}

= \frac{8}{3} - \frac{12 y}{3} + \frac{6 y ^{2}}{3} - \frac{y ^{3}}{3}

Divide the numbers:

\frac{12}{3} = 4

= \frac{8}{3} - 4 y + \frac{6 y ^{2}}{3} - \frac{y ^{3}}{3}

Divide the numbers:

\frac{6}{3} = 2

= \frac{8}{3} - 4 y + 2 y^{2} - \frac{y ^{3}}{3}

Rewrite in standard form

= - \frac{y ^{3}}{3} + 2 y^{2} - 4 y + \frac{8}{3}

\frac{s ^{2} + 1}{s ( s + 3 )}

(A + B) (B + A^{^{'}})

(- 1)^{n - 1} (\frac{1}{7})^{3 (n - 1)} (n - 1)!

n^{\frac{1}{2}} (1 + ln ∣ n ∣)

- \frac{t}{( t + x ) ( t - x )}

What is expand (((2-y)^3))/3 ?
The solution to expand (((2-y)^3))/3 is -(y^3)/3+2y^2-4y+8/3