What is expand-((4-3pi)a^3)/(18) ?

The solution to expand-((4-3pi)a^3)/(18) is -(2a^3)/9+(pia^3)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand-((4-3pi)a^3)/(18)

expand

- \frac{( 4 - 3 π ) a ^{3}}{18}

- \frac{2 a ^{3}}{9} + \frac{π a ^{3}}{6}

Solution steps

- \frac{( 4 - 3 π ) a ^{3}}{18}

Expand

(4 - 3 π) a^{3} : 4 a^{3} - 3 π a^{3}

= - \frac{4 a ^{3} - 3 π a ^{3}}{18}

Apply the fraction rule:

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

\frac{4 a ^{3} - 3 π a ^{3}}{18} = - (\frac{4 a ^{3}}{18}) - (- \frac{3 π a ^{3}}{18})

= - (\frac{4 a ^{3}}{18}) - (- \frac{3 π a ^{3}}{18})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{4 a ^{3}}{18} + \frac{3 π a ^{3}}{18}

Cancel

\frac{4 a ^{3}}{18} : \frac{2 a ^{3}}{9}

= - \frac{2 a ^{3}}{9} + \frac{3 π a ^{3}}{18}

Cancel

\frac{3 π a ^{3}}{18} : \frac{π a ^{3}}{6}

= - \frac{2 a ^{3}}{9} + \frac{π a ^{3}}{6}

(e^{10} - 1)^{2}

8 + (8 - cos (4 t)) U (t - 2 π)

\frac{lo g _{10} ( x )}{( 1 + lo g _{10} ( x ) ) ^{2}}

ux (u d x + x d u)

\frac{x + 3}{( x - 5 ) ( x ^{2} + 7 )}

What is expand-((4-3pi)a^3)/(18) ?
The solution to expand-((4-3pi)a^3)/(18) is -(2a^3)/9+(pia^3)/6