Derivative: Simplifying an Equation

[needOfHelpCMath]

How does this equation:

\(\displaystyle \dfrac{12x\sqrt{2x^3+3x+2}-\frac{\left(6x^2+3\right)^2}{2\sqrt{2x^3+3x+2}}}{2\left(2x^3+3x+2\right)}\)

becomes this equation

\(\displaystyle {12x^4+36x^2+48x-9}frac{4\left(2x^3+3x+2\right)^\frac{3}{2}}\)

 

[MarkFL]

\(\displaystyle \frac{12x\sqrt{2x^3+3x+2}-\dfrac{\left(6x^2+3\right)^2}{2\sqrt{2x^3+3x+2}}}{2\left(2x^3+3x+2\right)}\cdot\frac{2\sqrt{2x^3+3x+2}}{2\sqrt{2x^3+3x+2}}=\frac{24x(2x^3+3x+2)-(6x^2+3)^2}{4\left(2x^3+3x+2\right)^{\frac{3}{2}}}=\frac{48x^4+72x^2+48x-36x^4-36x^2-9}{4\left(2x^3+3x+2\right)^{\frac{3}{2}}}=\frac{12x^4+36x^2+48x-9}{4\left(2x^3+3x+2\right)^{\frac{3}{2}}}\)

 

[needOfHelpCMath]

\(\displaystyle \frac{12x\sqrt{2x^3+3x+2}-\dfrac{\left(6x^2+3\right)^2}{2\sqrt{2x^3+3x+2}}}{2\left(2x^3+3x+2\right)}\cdot\frac{2\sqrt{2x^3+3x+2}}{2\sqrt{2x^3+3x+2}}=\frac{24x(2x^3+3x+2)-(6x^2+3)^2}{4\left(2x^3+3x+2\right)^{\frac{3}{2}}}=\frac{48x^4+72x^2+48x-36x^4-36x^2-9}{4\left(2x^3+3x+2\right)^{\frac{3}{2}}}=\frac{12x^4+36x^2+48x-9}{4\left(2x^3+3x+2\right)^{\frac{3}{2}}}\)

ahh okay i see it now cancels out then foil for the left side then. thank you