What is expand ((-6+3y+4x)(18+3y-4x))/9 ?

The solution to expand ((-6+3y+4x)(18+3y-4x))/9 is y^2+4y-(16x^2)/9+(32x)/3-12

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expand ((-6+3y+4x)(18+3y-4x))/9

expand

\frac{( - 6 + 3 y + 4 x ) ( 18 + 3 y - 4 x )}{9}

y^{2} + 4 y - \frac{16 x ^{2}}{9} + \frac{32 x}{3} - 12

Solution steps

\frac{( - 6 + 3 y + 4 x ) ( 18 + 3 y - 4 x )}{9}

Expand

(- 6 + 3 y + 4 x) (18 + 3 y - 4 x) : 9 y^{2} + 36 y - 16 x^{2} + 96 x - 108

= \frac{9 y ^{2} + 36 y - 16 x ^{2} + 96 x - 108}{9}

Apply the fraction rule:

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

\frac{9 y ^{2} + 36 y - 16 x ^{2} + 96 x - 108}{9} = \frac{9 y ^{2}}{9} + \frac{36 y}{9} - \frac{16 x ^{2}}{9} + \frac{96 x}{9} - \frac{108}{9}

= \frac{9 y ^{2}}{9} + \frac{36 y}{9} - \frac{16 x ^{2}}{9} + \frac{96 x}{9} - \frac{108}{9}

\frac{9 y ^{2}}{9} = y^{2}

\frac{36 y}{9} = 4 y

\frac{96 x}{9} = \frac{32 x}{3}

\frac{108}{9} = 12

= y^{2} + 4 y - \frac{16 x ^{2}}{9} + \frac{32 x}{3} - 12

3 x^{2} y d x + (y + x^{3}) d x

3 x^{2} y d x + (y + x^{3}) d y

j (z - e^{T (j - 1)}) (z - e^{- T (j + 1)})

e^{2 x} (2 + e^{x})

\frac{8 x ( x + 4 )}{3 ( x - 4 )}

What is expand ((-6+3y+4x)(18+3y-4x))/9 ?
The solution to expand ((-6+3y+4x)(18+3y-4x))/9 is y^2+4y-(16x^2)/9+(32x)/3-12