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Question Number 55301 by Otchere Abdullai last updated on 20/Feb/19

$$findthefourthtermintheexpansion$$$$of{(\frac{\sqrt{x}}{y^2}-\frac{y}{\sqrt{x}})}^6$$

Commented by mr W last updated on 21/Feb/19

$${(\frac{\sqrt{x}}{y^2}-\frac{y}{\sqrt{x}})}^6$$$$={\left(\frac{\sqrt{x}}{y^2}\right)}^6{(1-\frac{y^3}{x})}^6$$$$=\frac{x^3}{y^{12}}{(1-\frac{y^3}{x})}^6$$$$4thterm=\frac{x^3}{y^{12}}\times C_3^6\times {(-\frac{y^3}{x})}^3$$$$=\frac{x^3}{y^{12}}\times 20\times (-\frac{y^9}{x^3})$$$$=-\frac{20}{y^3}$$

Commented by Otchere Abdullai last updated on 21/Feb/19

$$GodblessyouProfW$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Feb/19

$$r+1thterm\to {nc}_r{\left(a\right)}^{n-r}{\left(b\right)}^r$$$$so4thterm\to 6c_3{\left(\frac{\sqrt{x}}{y^2}\right)}^{6-3}{\left(\frac{-y}{\sqrt{x}}\right)}^3$$$$=\frac{6!}{3!\times 3!}\frac{{\left(x\right)}^{\frac{3}{2}}}{{\left(y\right)}^6}\times (-1)\times \frac{y^3}{{\left(x\right)}^{\frac{3}{2}}}$$$$=\frac{-20}{y^3}$$

Commented by Otchere Abdullai last updated on 21/Feb/19

$$thankyouprofTanmay$$

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