What is ((3x^3)/2+9y^2)((3x^3)/2-9y^2) ?

The solution to ((3x^3)/2+9y^2)((3x^3)/2-9y^2) is ((18y^2+3x^3)(-18y^2+3x^3))/4

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simplify ((3x^3)/2+9y^2)((3x^3)/2-9y^2)

simplify

(\frac{3 x ^{3}}{2} + 9 y^{2}) (\frac{3 x ^{3}}{2} - 9 y^{2})

\frac{( 18 y ^{2} + 3 x ^{3} ) ( - 18 y ^{2} + 3 x ^{3} )}{4}

Solution steps

(\frac{3 x ^{3}}{2} + 9 y^{2}) (\frac{3 x ^{3}}{2} - 9 y^{2})

Join

\frac{3 x ^{3}}{2} + 9 y^{2} : \frac{18 y ^{2} + 3 x ^{3}}{2}

= \frac{18 y ^{2} + 3 x ^{3}}{2} (\frac{3 x ^{3}}{2} - 9 y^{2})

Apply the fraction rule:

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

= \frac{( 18 y ^{2} + 3 x ^{3} ) ( \frac{3 x ^{3}}{2} - 9 y ^{2} )}{2}

Simplify

(18 y^{2} + 3 x^{3}) (\frac{3 x ^{3}}{2} - 9 y^{2}) : \frac{( 18 y ^{2} + 3 x ^{3} ) ( - 18 y ^{2} + 3 x ^{3} )}{2}

= \frac{\frac{( 18 y ^{2} + 3 x ^{3} ) ( - 18 y ^{2} + 3 x ^{3} )}{2}}{2}

Apply the fraction rule:

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

= \frac{( 18 y ^{2} + 3 x ^{3} ) ( - 18 y ^{2} + 3 x ^{3} )}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{( 18 y ^{2} + 3 x ^{3} ) ( - 18 y ^{2} + 3 x ^{3} )}{4}

(x + 1) (x - 4) (x - 2)

(\frac{A}{6} + 4 Z) (A - 5 Z)

(- x^{2} + 8 \circ 4 x + 8) (- 4)

z (- x_{1} + x_{2} - 4)^{2}

s (s + 1) (\frac{s}{2} + 1)

What is ((3x^3)/2+9y^2)((3x^3)/2-9y^2) ?
The solution to ((3x^3)/2+9y^2)((3x^3)/2-9y^2) is ((18y^2+3x^3)(-18y^2+3x^3))/4