$I\mathrm{.})49a^6+\mathrm{\square }+100c^2=(\mathrm{\square }+\mathrm{\square }{)}^2\backslash \; I.)\backslash quad49a^6+\backslash square+100c^2=(\backslash square+\backslash square)^2$

Wende die 1. Binomische Formel an:

$a^2+2ab+b^2=(a+b{)}^2\backslash hspace\{10mm\}a^2+2ab+b^2\backslash hspace\{3.7mm\}=\backslash hspace\{2mm\}\; (a+b)^2$

$49a^6+\boxed{{\displaystyle 140a^{3}c}}+100c^2={(\boxed{{\displaystyle 7a^3}}+\boxed{{\displaystyle 10c}})}^2\backslash quad49a^6+\backslash boxed\{140a^3c\}+100c^2=\backslash left(\backslash boxed\{7a^3\}+\backslash boxed\{10c\}\backslash right)^2$

$II\mathrm{.})(18ac+\mathrm{\square })\cdot (18ac-\mathrm{\square })=\mathrm{\square }-{\displaystyle \frac{1}{25}}{b}^2\backslash \; II.)\backslash quad(18ac+\backslash square)\backslash cdot\; (18ac-\backslash square)=\backslash square-\backslash dfrac\{1\}\{25\}b^2$

Wende die 3. Binomische Formel an:

$(a+b)\cdot (a-b)=a^2-b^2\backslash hspace\{12mm\}(a+b)\backslash hspace\{3mm\}\backslash cdot\backslash hspace\{3.5mm\}(a-b)\backslash hspace\{9mm\}=\backslash hspace\{10mm\}\; a^2-b^2$

$(18ac+\boxed{{\displaystyle \frac{1}{5}b}})\cdot (18ac-\boxed{{\displaystyle \frac{1}{5}b}})=\boxed{{\displaystyle 324a^2{c}^2}}-{\displaystyle \frac{1}{25}}{b}^2\backslash quad\backslash left(18ac+\backslash boxed\{\backslash dfrac\{1\}\{5\}b\}\backslash right)\backslash cdot\backslash left(18ac-\backslash boxed\{\backslash dfrac\{1\}\{5\}b\}\backslash right)=\backslash boxed\{324a^2c^2\}-\backslash dfrac\{1\}\{25\}b^2$