Jeśli mamy dany dwumian postaci:

${\left(a+b\right)}^n$  

to rozwinięcie tego dwumianu jest sumą składników postaci:

$\left(\begin{array}{c}n\\ k\end{array}\right)\cdot a^{n-k}\cdot b^k$  


natomiast gdy dany jest dwumian postaci:

${\left(a-b\right)}^n$  

to rozwinięcie tego dwumianu jest sumą składników postaci:

${\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right)\cdot a^{n-k}\cdot b^k$  

zatem:  

$a\text{)}{\left(2a^2+3a\right)}^7=\left(\begin{array}{c}7\\ k\end{array}\right)\cdot {\left(2a^2\right)}^{7-k}\cdot {\left(3a\right)}^k$  

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$b\text{)}{\left(3x-4y\right)}^9={\left(-1\right)}^k\cdot \left(\begin{array}{c}9\\ k\end{array}\right)\cdot {\left(3x\right)}^{9-k}\cdot {\left(4y\right)}^k$  

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$c\text{)}{\left(\sqrt{\frac{2}{3}a}+\sqrt{\frac{3}{2}x}\right)}^6=\left(\begin{array}{c}6\\ k\end{array}\right)\cdot {\left(\sqrt{\frac{2}{3}a}\right)}^{6-k}\cdot {\left(\sqrt{\frac{3}{2}x}\right)}^k$  

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$d\text{)}{\left(\sqrt{x}+\sqrt[3]{x^2}\right)}^{10}=\left(\begin{array}{c}10\\ k\end{array}\right)\cdot {\left(\sqrt{x}\right)}^{10-k}\cdot {\left(\sqrt[3]{x^2}\right)}^k$  

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$e\text{)}{\left(x^2-2x^{-1}\right)}^{30}={\left(-1\right)}^k\cdot \left(\begin{array}{c}30\\ k\end{array}\right)\cdot {\left(x^2\right)}^{30-k}\cdot {\left(2x^{-1}\right)}^k$  

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$f\text{)}{\left(\frac{3}{a}+\frac{2}{\sqrt{a}}\right)}^{16}=\left(\begin{array}{c}16\\ k\end{array}\right)\cdot {\left(\frac{3}{a}\right)}^{16-k}\cdot {\left(\frac{2}{\sqrt{a}}\right)}^k$