What is expand-(-l(-2l(3a-l^2)-a)+3a)/2 ?

The solution to expand-(-l(-2l(3a-l^2)-a)+3a)/2 is -3l^2a+l^4-(la)/2-(3a)/2

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expand-(-l(-2l(3a-l^2)-a)+3a)/2

expand

- \frac{- l ( - 2 l ( 3 a - l ^{2} ) - a ) + 3 a}{2}

- 3 l^{2} a + l^{4} - \frac{l a}{2} - \frac{3 a}{2}

Solution steps

- \frac{- l ( - 2 l ( 3 a - l ^{2} ) - a ) + 3 a}{2}

Expand

- 2 l (3 a - l^{2}) - a : - 6 l a + 2 l^{3} - a

= - \frac{3 a - l ( - 6 l a - a + 2 l ^{3} )}{2}

Expand

- l (- 6 l a + 2 l^{3} - a) + 3 a : 6 l^{2} a - 2 l^{4} + l a + 3 a

= - \frac{6 l ^{2} a + l a + 3 a - 2 l ^{4}}{2}

Apply the fraction rule:

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

\frac{6 l ^{2} a - 2 l ^{4} + l a + 3 a}{2} = - (\frac{6 l ^{2} a}{2}) - (- \frac{2 l ^{4}}{2}) - (\frac{l a}{2}) - (\frac{3 a}{2})

= - (\frac{6 l ^{2} a}{2}) - (- \frac{2 l ^{4}}{2}) - (\frac{l a}{2}) - (\frac{3 a}{2})

Remove parentheses:

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

= - \frac{6 l ^{2} a}{2} + \frac{2 l ^{4}}{2} - \frac{l a}{2} - \frac{3 a}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3 l^{2} a + \frac{2 l ^{4}}{2} - \frac{l a}{2} - \frac{3 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3 l^{2} a + l^{4} - \frac{l a}{2} - \frac{3 a}{2}

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

\frac{x ^{4} + 4 x ^{2} - 9}{4 x ( x ^{2} + 6 x - 9 )}

\frac{( 4 x ^{2} + 5 )}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

\frac{3}{x ^{2} - 2 x + 1} + \frac{2}{x ^{2} - 1}

sin (e^{x} + x) (1 + cos (x))^{14}

What is expand-(-l(-2l(3a-l^2)-a)+3a)/2 ?
The solution to expand-(-l(-2l(3a-l^2)-a)+3a)/2 is -3l^2a+l^4-(la)/2-(3a)/2