\displaystyle{32}{x}^{{-{{1}}}}{y} Explanation: Given, \displaystyle{\left[\frac{{{2}{x}^{{2}}{y}}}{{x}^{{3}}}\right]}^{{3}}.{\left[\frac{{y}^{{1}}}{{{2}{x}}}\right]}^{{-{{2}}}} \displaystyle\Rightarrow{\left[\frac{{{2}{y}}}{{x}}\right]}^{{3}}.{\left[\frac{{{2}{x}}}{{y}}\right]}^{{2}} ...

Konstantinos Michailidis May 26, 2016 It is \displaystyle\frac{{{3}{x}}}{{{6}{x}^{{2}}}}\times\frac{{{2}{y}-{6}}}{{{y}-{3}}}=\frac{{{3}{x}}}{{{2}{x}\cdot{3}{x}}}\times{\left(\frac{{{2}\cdot{\left({y}-{3}\right)}}}{{{y}-{3}}}\right)}=\frac{{\cancel{{{3}{x}}}}}{{{2}{x}\cdot\cancel{{{3}{x}}}}}\times{\left(\frac{{{2}\cdot\cancel{{{y}-{3}}}}}{\cancel{{{y}-{3}}}}\right)}=\frac{{1}}{{{2}{x}}}\times{2}=\frac{{1}}{{\cancel{{2}}\cdot{x}}}\cdot\cancel{{2}}=\frac{{1}}{{x}}

\displaystyle-{y}{\left({x}+{y}\right)} Explanation: \displaystyle{x}^{{2}}-{y}^{{2}}\ \text{ is a }\ \text{difference of squares} \displaystyle\text{and factors in general as} \displaystyle•{\left({x}\right)}{a}^{{2}}-{b}^{{2}}={\left({a}-{b}\right)}{\left({a}+{b}\right)} ...

See a solution process below: Explanation: First, rewrite the expression as: \displaystyle{\left(-\frac{{8}}{{21}}\times-\frac{{7}}{{16}}\right)}{\left({x}^{{2}}\times{x}\right)}{\left({y}^{{3}}\times{y}^{{2}}\right)}\Rightarrow ...

\displaystyle\frac{{{6}{x}}}{{{45}{y}^{{5}}}} Every step shown to aid understanding. As you become practised you will be able to skip steps. Explanation: Splitting the given expression down into ...

Assume that \text{char} k \neq 2. If x²+1 is irreducible over k one has that k[x,y]/(x²+1,y²+1) \simeq k(i)[y]/(y²+1). Since in k(i)[y] one has (y²+1) = (y+i)(y-i) and since (y+i) - (y-i) = 2i \in k(i)^\times ...