Răspuns:
Explicație pas cu pas:
[tex]T_{k+1}=C_{n}^{k}a^{n-k}b^{k},~unde~k+1=3,~adica~k=2,~n=6,~a=\frac{\sqrt[3]{a} }{\sqrt{b}},~b=\sqrt{ab} ,~deci\\ T_{2+1}=C_{6}^{2}*(\frac{\sqrt[3]{a} }{\sqrt{b}})^{6-2}*(\sqrt{ab})^{2}=\frac{6!}{2!*4!} *\frac{a\sqrt[3]{a} }{b^{2}} *ab=\frac{5*6}{2}*\frac{a^{2}\sqrt[3]{a} }{b} =15*\frac{a^{2}\sqrt[3]{a} }{b}[/tex]
