What is expand (WL(2X-L)^2)/8 ?

The solution to expand (WL(2X-L)^2)/8 is (WLX^2)/2-(WL^2X)/2+(WL^3)/8

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expand (WL(2X-L)^2)/8

expand

\frac{W L ( 2 X - L ) ^{2}}{8}

\frac{W L X ^{2}}{2} - \frac{W L ^{2} X}{2} + \frac{W L ^{3}}{8}

Solution steps

\frac{W L ( 2 X - L ) ^{2}}{8}

(2 X - L)^{2} = 4 X^{2} - 4 L X + L^{2}

= \frac{W L ( L ^{2} - 4 L X + 4 X ^{2} )}{8}

Expand

W L (4 X^{2} - 4 L X + L^{2}) : 4 W L X^{2} - 4 W L^{2} X + W L^{3}

= \frac{4 W L X ^{2} - 4 W L ^{2} X + W L ^{3}}{8}

Apply the fraction rule:

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

\frac{4 W L X ^{2} - 4 W L ^{2} X + W L ^{3}}{8} = \frac{4 W L X ^{2}}{8} - \frac{4 W L ^{2} X}{8} + \frac{W L ^{3}}{8}

= \frac{4 W L X ^{2}}{8} - \frac{4 W L ^{2} X}{8} + \frac{W L ^{3}}{8}

Cancel

\frac{4 W L X ^{2}}{8} : \frac{W L X ^{2}}{2}

= \frac{W L X ^{2}}{2} - \frac{4 W L ^{2} X}{8} + \frac{W L ^{3}}{8}

Cancel

\frac{4 W L ^{2} X}{8} : \frac{W L ^{2} X}{2}

= \frac{W L X ^{2}}{2} - \frac{W L ^{2} X}{2} + \frac{W L ^{3}}{8}

A (- 2, 1) B (3, 4) C (5, - 2)

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

\frac{a - b \cdot i}{1 + ( a + b \cdot i ) ^{2}}

\frac{25 ( 4 π - 3 3 )}{96}

\frac{d}{d x} (\frac{2 x y ^{2}}{( 1 + 2 x ^{2} ) ^{2}})

What is expand (WL(2X-L)^2)/8 ?
The solution to expand (WL(2X-L)^2)/8 is (WLX^2)/2-(WL^2X)/2+(WL^3)/8