${\displaystyle {\begin{aligned}(x-y)^{3}-\left(x^{3}-y^{3}\right)&=(x-y)^{3}-(x-y)(x^{2}+xy+y^{2})\\&=(x-y)\left[(x-y)^{2}-(x^{2}+xy+y^{2})\right]\\&=(x-y)(x^{2}-2xy+y^{2}-x^{2}-xy-y^{2})\\&=-3xy(x-y)\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {x^{2}-y^{2}}{x^{3}-y^{3}}}&={\frac {(x-y)(x+y)}{(x-y)(x^{2}+xy+y^{2})}}\\&={\frac {x+y}{x^{2}+xy+y^{2}}}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {x^{3}-1}{x^{3}-3x^{2}+3x-1}}&={\frac {(x-1)(x^{2}+x+1)}{(x-1)^{3}}}\\&={\frac {x^{2}+x+1}{(x-1)^{2}}}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {x^{3}+27}{x^{2}+3x}}&={\frac {(x+3)(x^{2}-3x+9)}{x(x+3)}}\\&={\frac {x^{2}-3x+9}{x}}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {(x^{2}+x+1)(x^{2}-x+1)}{x^{6}-1}}&={\frac {(x^{2}+x+1)(x^{2}-x+1)}{(x^{3}+1)(x^{3}-1)}}\\&={\frac {(x^{2}+x+1)(x^{2}-x+1)}{(x+1)(x^{2}-x+1)(x-1)(x^{2}+x+1)}}\\&={\frac {1}{(x+1)(x-1)}}\\&={\frac {1}{x^{2}-1}}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {x^{6}+2x^{3}y^{3}+y^{6}}{x^{2}+2xy+y^{2}}}&={\frac {(x^{3}+y^{3})^{2}}{(x+y)^{2}}}\\&={\frac {(x+y)^{2}(x^{2}-xy+y^{2})^{2}}{(x+y)^{2}}}\\&=(x^{2}-xy+y^{2})^{2}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {(x+1)^{4}-(x-1)^{4}}{8x^{5}+16x^{3}+8x}}&={\frac {\left[(x+1)^{2}-(x-1)^{2}\right]\left[(x+1)^{2}+(x-1)^{2}\right]}{8x(x^{4}+2x^{2}+1)}}\\&={\frac {\left[(x+1)-(x-1)\right]\left[(x+1)+(x-1)\right]\left[(x^{2}+2x+1)+(x^{2}-2x+1)\right]}{8x(x^{2}+1)^{2}}}\\&={\frac {2\times 2x(2x^{2}+2)}{8x(x^{2}+1)^{2}}}\\&={\frac {8x(x^{2}+1)}{8x(x^{2}+1)^{2}}}\\&={\frac {1}{x^{2}+1}}\end{aligned}}}$

Autre possibilité :

${\displaystyle {\begin{aligned}{\frac {(x+1)^{4}-(x-1)^{4}}{8x^{5}+16x^{3}+8x}}&={\frac {\left(x^{4}+4x^{3}+6x^{2}+4x+1\right)-\left(x^{4}-4x^{3}+6x^{2}-4x+1\right)}{8x(x^{4}+2x^{2}+1)}}\\&={\frac {8x^{3}+8x}{8x(x^{2}+1)^{2}}}\\&={\frac {8x(x^{2}+1)}{8x(x^{2}+1)^{2}}}\\&={\frac {1}{x^{2}+1}}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {x^{8}-y^{8}}{x^{6}-x^{4}y^{2}+x^{2}y^{4}-y^{6}}}&={\frac {(x^{4}+y^{4})(x^{4}-y^{4})}{x^{4}(x^{2}-y^{2})+y^{4}(x^{2}-y^{2})}}\\&={\frac {(x^{4}+y^{4})(x^{2}+y^{2})(x^{2}-y^{2})}{(x^{4}+y^{4})(x^{2}-y^{2})}}\\&=x^{2}+y^{2}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {x^{3}+1}{(x^{2}-1)(x^{2}-x+1)}}&={\frac {(x+1)(x^{2}-x+1)}{(x+1)(x-1)(x^{2}-x+1)}}\\&={\frac {1}{x-1}}\end{aligned}}}$

On trouve :

${\displaystyle 4abc}$

On trouve :

${\displaystyle b^{2}(a+3b)^{2}}$

On peut simplifier le calcul en posant :

${\displaystyle A=a^{2}+b^{2}+c^{2}}$

${\displaystyle B=bc+ca+ab}$

On a alors :

${\displaystyle {\begin{aligned}(a^{2}+b^{2}+c^{2}+bc+ca+ab)^{2}-(a+b+c)^{2}(a^{2}+b^{2}+c^{2})&=(a^{2}+b^{2}+c^{2}+bc+ca+ab)^{2}-(a^{2}+b^{2}+c^{2}+2ab+2ac+2bc)(a^{2}+b^{2}+c^{2})\\&=(A+B)^{2}-(A+2B)(A)\\&=A^{2}+2AB+B^{2}-A^{2}-2AB\\&=B^{2}\\&=(bc+ca+ab)^{2}\end{aligned}}}$

On a :

${\displaystyle (abc+bcd+cda+dab)^{2}=a^{2}b^{2}c^{2}+b^{2}c^{2}d^{2}+c^{2}d^{2}a^{2}+d^{2}a^{2}b^{2}+abcd(2ab+2ac+2ad+2bc+2bd+2cd)}$

${\displaystyle (bc-ad)(ca-bd)(ab-cd)=a^{2}b^{2}c^{2}+b^{2}c^{2}d^{2}+c^{2}d^{2}a^{2}+d^{2}a^{2}b^{2}-abcd\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}$

Par différence membre à membre, on obtient :

${\displaystyle {\begin{aligned}(abc+bcd+cda+dab)^{2}-(bc-ad)(ca-bd)(ab-cd)&=abcd(2ab+2ac+2ad+2bc+2bd+2cd)+abcd\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\\&=abcd\left(a^{2}+b^{2}+c^{2}+d^{2}+2ab+2ac+2ad+2bc+2bd+2cd\right)\\&=abcd(a+b+c+d)^{2}\end{aligned}}}$

${\displaystyle a^{3}-3a^{2}b+3ab^{2}-b^{3}=(a-b)^{3}}$

${\displaystyle {\begin{aligned}a^{3}-a^{2}b-ab^{2}+b^{3}&=a^{3}+b^{3}-a^{2}b-ab^{2}\\&=(a+b)(a^{2}-ab+b^{2})-ab(a+b)\\&=(a+b)(a^{2}-ab+b^{2}-ab)\\&=(a+b)(a^{2}-2ab+b^{2})\\&=(a+b)(a-b)^{2}\end{aligned}}}$

ou

${\displaystyle {\begin{aligned}a^{3}-a^{2}b-ab^{2}+b^{3}&=a^{2}(a-b)-b^{2}(a-b)\\&=(a^{2}-b^{2})(a-b)\\&=(a+b)(a-b)(a-b)\\&=(a+b)(a-b)^{2}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\sqrt {a^{2}+6a+9}}+{\sqrt {a^{2}-10a+25}}&={\sqrt {(a+3)^{2}}}+{\sqrt {(a-5)^{2}}}\\&=a+3+a-5\\&=2a-2\\&=2(a-1)\end{aligned}}}$

${\displaystyle {\begin{aligned}{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}{\sqrt {2-{\sqrt {2}}}}&={\sqrt {2}}{\sqrt {(2+{\sqrt {2}})(2-{\sqrt {2}})}}\\&={\sqrt {2}}{\sqrt {2^{2}-\left({\sqrt {2}}\right)^{2}}}\\&={\sqrt {2}}{\sqrt {4-2}}\\&={\sqrt {2}}{\sqrt {2}}\\&=2\end{aligned}}}$

${\displaystyle {\begin{aligned}({\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}})({\sqrt {2}}-{\sqrt {3}}+{\sqrt {5}})({\sqrt {5}}-{\sqrt {2}})&=(2-{\sqrt {6}}+{\sqrt {10}}+{\sqrt {6}}-3+{\sqrt {15}}+{\sqrt {10}}-{\sqrt {15}}+5)({\sqrt {5}}-{\sqrt {2}})\\&=(4+2{\sqrt {10}})({\sqrt {5}}-{\sqrt {2}})\\&=4{\sqrt {5}}-4{\sqrt {2}}+2{\sqrt {50}}-2{\sqrt {20}}\\&=4{\sqrt {5}}-4{\sqrt {2}}+10{\sqrt {2}}-4{\sqrt {5}}\\&=6{\sqrt {2}}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {3{\sqrt {8}}-2{\sqrt {12}}+{\sqrt {20}}}{3{\sqrt {18}}-2{\sqrt {27}}+{\sqrt {45}}}}&={\frac {6{\sqrt {2}}-4{\sqrt {3}}+2{\sqrt {5}}}{9{\sqrt {2}}-6{\sqrt {3}}+3{\sqrt {5}}}}\\&={\frac {2\left(3{\sqrt {2}}-2{\sqrt {3}}+{\sqrt {5}}\right)}{3\left(3{\sqrt {2}}-2{\sqrt {3}}+{\sqrt {5}}\right)}}\\&={\frac {2}{3}}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {3}{{\sqrt {5}}-{\sqrt {2}}}}+{\frac {4}{{\sqrt {6}}+{\sqrt {2}}}}&={\frac {3\left({\sqrt {5}}+{\sqrt {2}}\right)}{\left({\sqrt {5}}-{\sqrt {2}}\right)\left({\sqrt {5}}+{\sqrt {2}}\right)}}+{\frac {4\left({\sqrt {6}}-{\sqrt {2}}\right)}{\left({\sqrt {6}}+{\sqrt {2}}\right)\left({\sqrt {6}}-{\sqrt {2}}\right)}}\\&={\frac {3\left({\sqrt {5}}+{\sqrt {2}}\right)}{5-2}}+{\frac {4\left({\sqrt {6}}-{\sqrt {2}}\right)}{6-2}}\\&={\frac {3\left({\sqrt {5}}+{\sqrt {2}}\right)}{3}}+{\frac {4\left({\sqrt {6}}-{\sqrt {2}}\right)}{4}}\\&={\sqrt {5}}+{\sqrt {2}}+{\sqrt {6}}-{\sqrt {2}}\\&={\sqrt {5}}+{\sqrt {6}}\end{aligned}}}$