Multiplying & Dividing Rational Expressions

\(\frac{5y^2}{3} \cdot \frac{9x}{10y}\)

The terms are already factored.

\(\frac{5y^2}{3} \cdot \frac{9x}{10y} = \frac{5y \cdot y}{3} \cdot \frac{9x}{10 \cdot y}\)

We can cancel one \(y\) from numerator and denominator:

\(\frac{5y \cdot \color{red}{y}}{3} \cdot \frac{9x}{10 \cdot \color{red}{y}} = \frac{5y}{3} \cdot \frac{9x}{10}\)

\(\frac{5y}{3} \cdot \frac{9x}{10} = \frac{5y \cdot 9x}{3 \cdot 10} = \frac{45xy}{30} = \frac{3xy}{2}\)

\(\frac{2x^3y}{3z^4} \cdot \frac{6xz^5}{10y^5}\)

\(\frac{2x^3y}{3z^4} \cdot \frac{6xz^5}{10y^5} = \frac{2x^3y}{3z^4} \cdot \frac{6xz^5}{2 \cdot 5 \cdot y^5}\)

We can cancel a factor of 2:

\(\frac{\color{red}{2}x^3y}{3z^4} \cdot \frac{6xz^5}{\color{red}{2} \cdot 5 \cdot y^5} = \frac{x^3y}{3z^4} \cdot \frac{6xz^5}{5 \cdot y^5}\)

No more common factors to cancel.

\(\frac{x^3y}{3z^4} \cdot \frac{6xz^5}{5 \cdot y^5} = \frac{x^3y \cdot 6xz^5}{3z^4 \cdot 5 \cdot y^5} = \frac{6x^4yz^5}{15z^4y^5} = \frac{6x^4z^5}{15z^4y^4} = \frac{6x^4z}{15y^4}\)

Simplifying further:

\(\frac{6x^4z}{15y^4} = \frac{2x^4z}{5y^4}\)

\(\frac{9y^2}{8} \cdot \frac{32x}{27y}\)

\(\frac{9y^2}{8} \cdot \frac{32x}{27y} = \frac{3^2 \cdot y^2}{2^3} \cdot \frac{2^5 \cdot x}{3^3 \cdot y}\)

We can cancel \(y\) once from numerator and denominator:

\(\frac{3^2 \cdot y \cdot \color{red}{y}}{2^3} \cdot \frac{2^5 \cdot x}{3^3 \cdot \color{red}{y}} = \frac{3^2 \cdot y}{2^3} \cdot \frac{2^5 \cdot x}{3^3}\)

\(\frac{3^2 \cdot y}{2^3} \cdot \frac{2^5 \cdot x}{3^3} = \frac{3^2 \cdot 2^5 \cdot y \cdot x}{2^3 \cdot 3^3} = \frac{9 \cdot 32 \cdot xy}{8 \cdot 27} = \frac{288xy}{216} = \frac{4xy}{3}\)

\(\frac{2x^2y}{3z^3} \cdot \frac{12xz^4}{6y^3}\)

\(\frac{2x^2y}{3z^3} \cdot \frac{12xz^4}{6y^3} = \frac{2x^2y}{3z^3} \cdot \frac{2 \cdot 6 \cdot xz^4}{6y^3} = \frac{2x^2y}{3z^3} \cdot \frac{2 \cdot 6 \cdot xz^4}{6y^3}\)

We can cancel 6 in the second fraction:

\(\frac{2x^2y}{3z^3} \cdot \frac{2 \cdot \color{red}{6} \cdot xz^4}{\color{red}{6}y^3} = \frac{2x^2y}{3z^3} \cdot \frac{2xz^4}{y^3}\)

\(\frac{2x^2y}{3z^3} \cdot \frac{2xz^4}{y^3} = \frac{2x^2y \cdot 2xz^4}{3z^3 \cdot y^3} = \frac{4x^3yz^4}{3z^3y^3} = \frac{4x^3z^4}{3z^3y^2} = \frac{4x^3z}{3y^2}\)

\(\frac{3y^2}{5} \cdot \frac{10x}{15y}\)

\(\frac{3y^2}{5} \cdot \frac{10x}{15y} = \frac{3y^2}{5} \cdot \frac{2 \cdot 5 \cdot x}{3 \cdot 5 \cdot y}\)

We can cancel 5 and 3 and one y:

\(\frac{\color{red}{3}y \cdot \color{red}{y}}{\color{red}{5}} \cdot \frac{2 \cdot \color{red}{5} \cdot x}{\color{red}{3} \cdot 5 \cdot \color{red}{y}} = \frac{y}{1} \cdot \frac{2x}{5}\)

\(\frac{y}{1} \cdot \frac{2x}{5} = \frac{y \cdot 2x}{1 \cdot 5} = \frac{2xy}{5}\)

\(\frac{4x^2y}{2z^2} \cdot \frac{6xz^3}{20y^4}\)

\(\frac{4x^2y}{2z^2} \cdot \frac{6xz^3}{20y^4} = \frac{2^2 \cdot x^2y}{2 \cdot z^2} \cdot \frac{6xz^3}{2^2 \cdot 5 \cdot y^4}\)

We can cancel 2 from the first fraction and 2² from the second:

\(\frac{\color{red}{2^2} \cdot x^2y}{\color{red}{2} \cdot z^2} \cdot \frac{6xz^3}{\color{red}{2^2} \cdot 5 \cdot y^4} = \frac{2 \cdot x^2y}{z^2} \cdot \frac{6xz^3}{5 \cdot y^4}\)

\(\frac{2 \cdot x^2y}{z^2} \cdot \frac{6xz^3}{5 \cdot y^4} = \frac{2 \cdot 6 \cdot x^2y \cdot xz^3}{z^2 \cdot 5 \cdot y^4} = \frac{12x^3yz^3}{5z^2y^4} = \frac{12x^3z}{5y^3}\)

\(\frac{x+4}{3x+4y} \cdot \frac{9x^2-16y^2}{2x^2+3x-20}\)

\(x+4\) is already factored
\(3x+4y = 3x+4y\) is already factored
\(9x^2-16y^2 = (3x)^2-(4y)^2 = (3x+4y)(3x-4y)\)
\(2x^2+3x-20 = 2x^2+8x-5x-20 = 2x(x+4)-(5)(x+4) = (x+4)(2x-5)\)

\(\frac{\color{red}{x+4}}{3x+4y} \cdot \frac{(3x+4y)(3x-4y)}{(2x-5)\color{red}{(x+4)}} = \frac{1}{3x+4y} \cdot \frac{(3x+4y)(3x-4y)}{2x-5}\)

We can also cancel \(3x+4y\):

\(\frac{1}{\color{red}{3x+4y}} \cdot \frac{\color{red}{(3x+4y)}(3x-4y)}{2x-5} = \frac{3x-4y}{2x-5}\)

\(\frac{a^2-10a+21}{a-7} \cdot \frac{a^2+a-12}{(a-3)^2}\)

\(a^2-10a+21 = (a-7)(a-3)\)
\(a-7\) is already factored
\(a^2+a-12 = (a+4)(a-3)\)
\((a-3)^2 = (a-3)(a-3)\)

\(\frac{(a-7)\color{red}{(a-3)}}{\color{red}{a-7}} \cdot \frac{(a+4)\color{red}{(a-3)}}{\color{red}{(a-3)}(a-3)} = \frac{a-3}{1} \cdot \frac{a+4}{a-3} = \frac{\color{red}{a-3} \cdot (a+4)}{\color{red}{a-3}} = a+4\)

\(\frac{x+1}{3x+y} \cdot \frac{9x^2-y^2}{2x^2+3x+1}\)

\(x+1\) is already factored
\(3x+y\) is already factored
\(9x^2-y^2 = (3x)^2-y^2 = (3x+y)(3x-y)\)
\(2x^2+3x+1 = 2x^2+2x+x+1 = 2x(x+1)+(x+1) = (x+1)(2x+1)\)

\(\frac{\color{red}{x+1}}{3x+y} \cdot \frac{(3x+y)(3x-y)}{(2x+1)\color{red}{(x+1)}} = \frac{1}{3x+y} \cdot \frac{(3x+y)(3x-y)}{2x+1}\)

We can also cancel \(3x+y\):

\(\frac{1}{\color{red}{3x+y}} \cdot \frac{\color{red}{(3x+y)}(3x-y)}{2x+1} = \frac{3x-y}{2x+1}\)

\(\frac{a^2-6a+9}{a-3} \cdot \frac{a^2+3a-18}{(a-3)^2}\)

\(a^2-6a+9 = (a-3)^2\)
\(a-3\) is already factored
\(a^2+3a-18 = (a+6)(a-3)\)
\((a-3)^2 = (a-3)(a-3)\)

\(\frac{(a-3)\color{red}{(a-3)}}{\color{red}{a-3}} \cdot \frac{(a+6)\color{red}{(a-3)}}{\color{red}{(a-3)}(a-3)} = \frac{a-3}{1} \cdot \frac{a+6}{a-3} = \frac{\color{red}{a-3} \cdot (a+6)}{\color{red}{a-3}} = a+6\)

\(\frac{5x^2}{3t^2} \cdot \frac{9t^8}{25x}\)

\(\frac{5x^2}{3t^2} \cdot \frac{9t^8}{25x} = \frac{5x^2}{3t^2} \cdot \frac{3^2 \cdot t^8}{5^2 \cdot x}\)

\(\frac{\color{red}{5}x^2}{3t^2} \cdot \frac{3^2 \cdot t^8}{\color{red}{5^2} \cdot x} = \frac{x^2}{3t^2} \cdot \frac{9 \cdot t^8}{5 \cdot x}\)

We can also cancel one power of \(x\):

\(\frac{\color{red}{x} \cdot x}{3t^2} \cdot \frac{9 \cdot t^8}{5 \cdot \color{red}{x}} = \frac{x}{3t^2} \cdot \frac{9t^8}{5}\)

\(\frac{x}{3t^2} \cdot \frac{9t^8}{5} = \frac{x \cdot 9t^8}{3t^2 \cdot 5} = \frac{9xt^8}{15t^2} = \frac{9xt^6}{15}\)

Simplifying further:

\(\frac{9xt^6}{15} = \frac{3xt^6}{5}\)

\(\frac{7a^3}{10b^7} \cdot \frac{5b^3}{3a}\)

\(\frac{7a^3}{10b^7} \cdot \frac{5b^3}{3a} = \frac{7a^3}{2 \cdot 5 \cdot b^7} \cdot \frac{5b^3}{3a}\)

\(\frac{7a^3}{2 \cdot \color{red}{5} \cdot b^7} \cdot \frac{\color{red}{5}b^3}{3a} = \frac{7a^3}{2 \cdot b^7} \cdot \frac{b^3}{3a}\)

We can also cancel one power of \(a\):

\(\frac{7a^2 \cdot \color{red}{a}}{2 \cdot b^7} \cdot \frac{b^3}{3 \cdot \color{red}{a}} = \frac{7a^2}{2 \cdot b^7} \cdot \frac{b^3}{3}\)

\(\frac{7a^2}{2 \cdot b^7} \cdot \frac{b^3}{3} = \frac{7a^2 \cdot b^3}{2 \cdot b^7 \cdot 3} = \frac{7a^2b^3}{6b^7} = \frac{7a^2}{6b^4}\)

\(\frac{3x-6}{5x} \cdot \frac{x^3}{5x-10}\)

\(3x-6 = 3(x-2)\)
\(5x = 5x\) is already factored
\(x^3 = x^3\) is already factored
\(5x-10 = 5(x-2)\)

\(\frac{3\color{red}{(x-2)}}{5x} \cdot \frac{x^3}{5\color{red}{(x-2)}} = \frac{3}{5x} \cdot \frac{x^3}{5} = \frac{3 \cdot x^3}{5x \cdot 5} = \frac{3x^3}{25x} = \frac{3x^2}{25}\)

\(\frac{5t^3}{4t-8} \cdot \frac{6t-12}{10t}\)

\(5t^3 = 5t^3\) is already factored
\(4t-8 = 4(t-2)\)
\(6t-12 = 6(t-2)\)
\(10t = 10t\) is already factored

\(\frac{5t^3}{4\color{red}{(t-2)}} \cdot \frac{6\color{red}{(t-2)}}{10t} = \frac{5t^3}{4} \cdot \frac{6}{10t} = \frac{5t^3}{4} \cdot \frac{6}{10t}\)

We can also simplify \(\frac{6}{10} = \frac{3}{5}\):

\(\frac{5t^3}{4} \cdot \frac{3}{5t} = \frac{5t^3 \cdot 3}{4 \cdot 5t} = \frac{15t^3}{20t} = \frac{15t^2}{20} = \frac{3t^2}{4}\)

\(\frac{y^2-16}{2y+6} \cdot \frac{y+3}{y-4}\)

\(y^2-16 = (y+4)(y-4)\)
\(2y+6 = 2(y+3)\)
\(y+3\) is already factored
\(y-4\) is already factored

\(\frac{(y+4)\color{red}{(y-4)}}{2\color{red}{(y+3)}} \cdot \frac{\color{red}{(y+3)}}{\color{red}{(y-4)}} = \frac{(y+4)}{2}\)

\(\frac{m^2-n^2}{4m+4n} \cdot \frac{m+n}{m-n}\)

\(m^2-n^2 = (m+n)(m-n)\)
\(4m+4n = 4(m+n)\)
\(m+n\) is already factored
\(m-n\) is already factored

\(\frac{\color{red}{(m+n)}(m-n)}{4\color{red}{(m+n)}} \cdot \frac{\color{red}{(m+n)}}{\color{red}{(m-n)}} = \frac{m-n}{4} \cdot \frac{1}{1} = \frac{m-n}{4}\)

\(\frac{x^2-16}{x^2} \cdot \frac{x^2-4x}{x^2-x-12}\)

\(x^2-16 = (x+4)(x-4)\)
\(x^2 = x^2\) is already factored
\(x^2-4x = x(x-4)\)
\(x^2-x-12 = (x+3)(x-4)\)

\(\frac{(x+4)\color{red}{(x-4)}}{x^2} \cdot \frac{x\color{red}{(x-4)}}{(x+3)\color{red}{(x-4)}} = \frac{(x+4)}{x^2} \cdot \frac{x}{(x+3)}\)

\(\frac{(x+4)}{x^2} \cdot \frac{x}{(x+3)} = \frac{(x+4) \cdot x}{x^2 \cdot (x+3)} = \frac{x(x+4)}{x^2(x+3)} = \frac{x+4}{x(x+3)}\)

\(\frac{y^2+10y+25}{y^2-9} \cdot \frac{y^2+3y}{y+5}\)

\(y^2+10y+25 = (y+5)^2\)
\(y^2-9 = (y+3)(y-3)\)
\(y^2+3y = y(y+3)\)
\(y+5\) is already factored

\(\frac{\color{red}{(y+5)}(y+5)}{(y+3)(y-3)} \cdot \frac{y\color{red}{(y+3)}}{\color{red}{(y+5)}} = \frac{(y+5)}{(y-3)} \cdot \frac{y}{1} = \frac{y(y+5)}{y-3}\)

\(\frac{6-2t}{t^2+4t+4} \cdot \frac{t^3+2t^2}{t^8-9t^6}\)

\(6-2t = 2(3-t)\)
\(t^2+4t+4 = (t+2)^2\)
\(t^3+2t^2 = t^2(t+2)\)
\(t^8-9t^6 = t^6(t^2-9) = t^6(t-3)(t+3)\)

\(\frac{2(3-t)}{(t+2)^2} \cdot \frac{t^2\color{red}{(t+2)}}{t^6(t-3)(t+3)}\)

To handle (3-t), we can rewrite it as -(t-3):

\(\frac{2(-(t-3))}{(t+2)^2} \cdot \frac{t^2\color{red}{(t+2)}}{t^6\color{red}{(t-3)}(t+3)} = \frac{-2(t-3)}{(t+2)(t+2)} \cdot \frac{t^2}{t^6(t+3)}\)

Canceling the common factor (t-3):

\(\frac{-2\color{red}{(t-3)}}{(t+2)^2} \cdot \frac{t^2}{t^6\color{red}{(t-3)}(t+3)} = \frac{-2}{(t+2)^2} \cdot \frac{t^2}{t^6(t+3)}\)

\(\frac{-2}{(t+2)^2} \cdot \frac{t^2}{t^6(t+3)} = \frac{-2 \cdot t^2}{(t+2)^2 \cdot t^6(t+3)} = \frac{-2t^2}{t^6(t+2)^2(t+3)} = \frac{-2}{t^4(t+2)^2(t+3)}\)

\(\frac{x^2-6x+9}{12-4x} \cdot \frac{x^6-9x^4}{x^3-3x^2}\)

\(x^2-6x+9 = (x-3)^2\)
\(12-4x = 4(3-x) = -4(x-3)\)
\(x^6-9x^4 = x^4(x^2-9) = x^4(x+3)(x-3)\)
\(x^3-3x^2 = x^2(x-3)\)

\(\frac{(x-3)^2}{-4(x-3)} \cdot \frac{x^4(x+3)\color{red}{(x-3)}}{x^2\color{red}{(x-3)}} = \frac{(x-3)}{-4} \cdot \frac{x^4(x+3)}{x^2}\)

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

\(\frac{-x^2(x-3)(x+3)}{4} = \frac{-x^2(x^2-9)}{4}\)

\(\frac{x^2-2x-35}{2x^3-3x^2} \cdot \frac{4x^3-9x}{7x-49}\)

\(x^2-2x-35 = (x+5)(x-7)\)
\(2x^3-3x^2 = x^2(2x-3)\)
\(4x^3-9x = x(4x^2-9) = x(2x+3)(2x-3)\)
\(7x-49 = 7(x-7)\)

\(\frac{(x+5)\color{red}{(x-7)}}{x^2(2x-3)} \cdot \frac{x(2x+3)\color{red}{(2x-3)}}{7\color{red}{(x-7)}} = \frac{(x+5)}{x^2\color{red}{(2x-3)}} \cdot \frac{x(2x+3)}{\color{red}{7}} = \frac{(x+5)(2x+3)}{7x}\)

\(\frac{y^2-10y+9}{y^2-1} \cdot \frac{y+4}{y^2-5y-36}\)

\(y^2-10y+9 = (y-9)(y-1)\)
\(y^2-1 = (y+1)(y-1)\)
\(y+4\) is already factored
\(y^2-5y-36 = (y+4)(y-9)\)

\(\frac{(y-9)\color{red}{(y-1)}}{(y+1)\color{red}{(y-1)}} \cdot \frac{\color{red}{(y+4)}}{\color{red}{(y+4)}(y-9)} = \frac{\color{red}{(y-9)}}{(y+1)} \cdot \frac{1}{\color{red}{(y-9)}} = \frac{1}{(y+1)}\)

\(\frac{c^3+8}{c^5-4c^3} \cdot \frac{c^6-4c^5+4c^4}{c^2-2c+4}\)

\(c^3+8 = c^3+2^3 = (c+2)(c^2-2c+4)\)
\(c^5-4c^3 = c^3(c^2-4) = c^3(c+2)(c-2)\)
\(c^6-4c^5+4c^4 = c^4(c^2-4c+4) = c^4(c-2)^2\)
\(c^2-2c+4\) is already factored

\(\frac{(c+2)\color{red}{(c^2-2c+4)}}{c^3(c+2)(c-2)} \cdot \frac{c^4(c-2)^2}{\color{red}{(c^2-2c+4)}} = \frac{\color{red}{(c+2)}}{c^3\color{red}{(c+2)}(c-2)} \cdot \frac{c^4(c-2)^2}{1}\)

\(\frac{1}{c^3(c-2)} \cdot \frac{c^4(c-2)^2}{1} = \frac{c^4(c-2)^2}{c^3(c-2)} = \frac{c^4(c-2)}{c^3} = \frac{c \cdot (c-2)}{1} = c(c-2)\)

\(\frac{x^3-27}{x^4-9x^2} \cdot \frac{x^5-6x^4+9x^3}{x^2+3x+9}\)

\(x^3-27 = (x-3)(x^2+3x+9)\)
\(x^4-9x^2 = x^2(x^2-9) = x^2(x+3)(x-3)\)
\(x^5-6x^4+9x^3 = x^3(x^2-6x+9) = x^3(x-3)^2\)
\(x^2+3x+9\) is already factored

\(\frac{(x-3)\color{red}{(x^2+3x+9)}}{x^2(x+3)(x-3)} \cdot \frac{x^3(x-3)^2}{\color{red}{(x^2+3x+9)}} = \frac{\color{red}{(x-3)}}{x^2(x+3)\color{red}{(x-3)}} \cdot \frac{x^3(x-3)^2}{1}\)

\(\frac{1}{x^2(x+3)} \cdot \frac{x^3(x-3)^2}{1} = \frac{x^3(x-3)^2}{x^2(x+3)} = \frac{x(x-3)^2}{(x+3)}\)

\(\frac{a^3-b^3}{3a^2+9ab+6b^2} \cdot \frac{a^2+2ab+b^2}{a^2-b^2}\)

\(a^3-b^3 = (a-b)(a^2+ab+b^2)\)
\(3a^2+9ab+6b^2 = 3(a^2+3ab+2b^2) = 3(a+2b)(a+b)\)
\(a^2+2ab+b^2 = (a+b)^2\)
\(a^2-b^2 = (a+b)(a-b)\)

\(\frac{(a-b)(a^2+ab+b^2)}{3(a+2b)\color{red}{(a+b)}} \cdot \frac{\color{red}{(a+b)}^2}{\color{red}{(a+b)}(a-b)} = \frac{(a^2+ab+b^2)}{3(a+2b)} \cdot \frac{(a+b)}{1}\)

\(\frac{(a^2+ab+b^2)(a+b)}{3(a+2b)} = \frac{(a^2+ab+b^2)(a+b)}{3(a+2b)}\)

\(\frac{x^3+y^3}{x^2+2xy-3y^2} \cdot \frac{x^2-y^2}{3x^2+6xy+3y^2}\)

\(x^3+y^3 = (x+y)(x^2-xy+y^2)\)
\(x^2+2xy-3y^2 = (x+3y)(x-y)\)
\(x^2-y^2 = (x+y)(x-y)\)
\(3x^2+6xy+3y^2 = 3(x^2+2xy+y^2) = 3(x+y)^2\)

\(\frac{\color{red}{(x+y)}(x^2-xy+y^2)}{(x+3y)\color{red}{(x-y)}} \cdot \frac{\color{red}{(x+y)}\color{red}{(x-y)}}{3\color{red}{(x+y)}^2} = \frac{(x^2-xy+y^2)}{(x+3y)} \cdot \frac{1}{3(x+y)}\)

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

\(\frac{4x^2-9y^2}{8x^3-27y^3} \cdot \frac{4x^2+6xy+9y^2}{4x^2+12xy+9y^2}\)

\(4x^2-9y^2 = (2x+3y)(2x-3y)\)
\(8x^3-27y^3 = (2x-3y)(4x^2+6xy+9y^2)\)
\(4x^2+6xy+9y^2 = (2x+3y)^2\)
\(4x^2+12xy+9y^2 = (2x+3y)^2\)

We notice that \(4x^2+12xy+9y^2\) factors differently from what was given. Let's double-check:

\(4x^2+12xy+9y^2 = 4x^2+12xy+9y^2 = (2x)^2+2(2x)(3y)+(3y)^2\)

This is \((2x+3y)^2\). Let's continue with the correct factorization:

\(\frac{\color{red}{(2x+3y)}(2x-3y)}{\color{red}{(2x-3y)}(4x^2+6xy+9y^2)} \cdot \frac{\color{red}{(2x+3y)}^2}{\color{red}{(2x+3y)}^2} = \frac{(2x+3y)}{(4x^2+6xy+9y^2)} = \frac{1}{(2x+3y)}\)

\(\frac{3x^2-3y^2}{27x^3-8y^3} \cdot \frac{6x^2+5xy-6y^2}{6x^2+12xy+6y^2}\)

\(3x^2-3y^2 = 3(x^2-y^2) = 3(x+y)(x-y)\)
\(27x^3-8y^3 = (3x-2y)(9x^2+6xy+4y^2)\)
\(6x^2+5xy-6y^2 = (3x+6y)(2x-y)\)
\(6x^2+12xy+6y^2 = 6(x^2+2xy+y^2) = 6(x+y)^2\)

\(\frac{3\color{red}{(x+y)}(x-y)}{(3x-2y)(9x^2+6xy+4y^2)} \cdot \frac{(3x+6y)(2x-y)}{6\color{red}{(x+y)}^2}\)

Let's factor further to find common terms:

\((2x-y)\) and \((x-y)\) have a common factor of \((x-y)\) if we write \((2x-y) = 2(x-\frac{y}{2})\).

Similarly, \((3x+6y) = 3(x+2y)\) and \((3x-2y)\) don't have obvious common factors.

Let's continue with our cancellation:

\(\frac{3(x-y)}{(3x-2y)(9x^2+6xy+4y^2)} \cdot \frac{(3x+6y)(2x-y)}{6(x+y)}\)

Without more common factors to cancel, we have:

\(\frac{3(x-y)(3x+6y)(2x-y)}{6(x+y)(3x-2y)(9x^2+6xy+4y^2)}\)

\(\frac{3(x-y)(3x+6y)(2x-y)}{6(x+y)(3x-2y)(9x^2+6xy+4y^2)} = \frac{(x-y)(3x+6y)(2x-y)}{2(x+y)(3x-2y)(9x^2+6xy+4y^2)}\)

\(\frac{(x-y)(3(x+2y))(2x-y)}{2(x+y)(3x-2y)(9x^2+6xy+4y^2)}\)