What is expand ((b^2+ab+bc+ac)(a+c))/(3abc) ?

The solution to expand ((b^2+ab+bc+ac)(a+c))/(3abc) is b/(3c)+b/(3a)+a/(3c)+2/3+c/(3a)+a/(3b)+c/(3b)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ((b^2+ab+bc+ac)(a+c))/(3abc)

expand

\frac{( b ^{2} + ab + b c + a c ) ( a + c )}{3 ab c}

\frac{b}{3 c} + \frac{b}{3 a} + \frac{a}{3 c} + \frac{2}{3} + \frac{c}{3 a} + \frac{a}{3 b} + \frac{c}{3 b}

Solution steps

\frac{( b ^{2} + ab + b c + a c ) ( a + c )}{3 ab c}

Expand

(b^{2} + ab + b c + a c) (a + c) : b^{2} a + b^{2} c + b a^{2} + 2 ba c + b c^{2} + a^{2} c + a c^{2}

= \frac{b ^{2} a + b ^{2} c + b a ^{2} + 2 ba c + b c ^{2} + a ^{2} c + a c ^{2}}{3 ba c}

Apply the fraction rule:

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

\frac{b ^{2} a + b ^{2} c + b a ^{2} + 2 ba c + b c ^{2} + a ^{2} c + a c ^{2}}{3 ab c} = \frac{b ^{2} a}{3 ab c} + \frac{b ^{2} c}{3 ab c} + \frac{b a ^{2}}{3 ab c} + \frac{2 ba c}{3 ab c} + \frac{b c ^{2}}{3 ab c} + \frac{a ^{2} c}{3 ab c} + \frac{a c ^{2}}{3 ab c}

= \frac{b ^{2} a}{3 ba c} + \frac{b ^{2} c}{3 ba c} + \frac{b a ^{2}}{3 ba c} + \frac{2 ba c}{3 ba c} + \frac{b c ^{2}}{3 ba c} + \frac{a ^{2} c}{3 ba c} + \frac{a c ^{2}}{3 ba c}

\frac{b ^{2} a}{3 ab c} = \frac{b}{3 c}

\frac{b ^{2} c}{3 ab c} = \frac{b}{3 a}

\frac{b a ^{2}}{3 ab c} = \frac{a}{3 c}

\frac{2 ba c}{3 ab c} = \frac{2}{3}

\frac{b c ^{2}}{3 ab c} = \frac{c}{3 a}

\frac{a ^{2} c}{3 ab c} = \frac{a}{3 b}

\frac{a c ^{2}}{3 ab c} = \frac{c}{3 b}

= \frac{b}{3 c} + \frac{b}{3 a} + \frac{a}{3 c} + \frac{2}{3} + \frac{c}{3 a} + \frac{a}{3 b} + \frac{c}{3 b}

\frac{x ^{2} - 4}{( x - 1 ) ^{2}}

\frac{( a ^{2} + 4 ) ^{2}}{4 a}

A (2, 4) B (- 8, - 12)

\frac{- 2 e ^{- x^{2}} x ^{3} - 4 e ^{- x^{2}}}{( x ^{2} + 1 ) ^{e}}

\frac{x}{( x + 2 ) ( x - 1 )}

What is expand ((b^2+ab+bc+ac)(a+c))/(3abc) ?
The solution to expand ((b^2+ab+bc+ac)(a+c))/(3abc) is b/(3c)+b/(3a)+a/(3c)+2/3+c/(3a)+a/(3b)+c/(3b)