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Question Number 88882 by Zainal Arifin last updated on 13/Apr/20

$$\mathrm{The}\mathrm{coefficient}\mathrm{of}{x}^3\mathrm{in}{(\sqrt{x^5}+\frac{3}{\sqrt{x^3}})}^6\mathrm{is}$$

Commented by mathmax by abdo last updated on 13/Apr/20

$$wehave{(\sqrt{x^5}+\frac{3}{\sqrt{x^3}})}^6={(x^{\frac{1}{5}}+3x^{-\frac{3}{2}})}^6$$$$=\sum _{k=0}^6{C}_6^k{\left(3x^{-\frac{3}{2}}\right)}^k{\left(x^{\frac{5}{2}}\right)}^{6-k}$$$$=\sum _{k=0}^6{C}_6^k{3}^k{x}^{-\frac{3k}{2}}{x}^{\frac{30-5k}{2}}=\sum _{k=0}^6{C}_6^k{3}^k{x}^{\frac{-3k+30-5k}{2}}$$$$wegetthecoefficientof{x}^{3}when\frac{-8k+30}{2}=3\Rightarrow$$$$-8k+30=6\Rightarrow 8k=24\Rightarrow k=3andthecoefficientis$$$$a=C_6^3{3}^3=27\times \frac{6!}{3!\times 3!}=\frac{27.6.5.4.3!}{3!.3!}=27.5.4=27\times 20=540$$

Answered by TANMAY PANACEA. last updated on 13/Apr/20

$$let(r+1)thtermcontains{x}^3$$$$6C_r.{\left(\sqrt{x^5}\right)}^{6-r}.{\left(\frac{3}{\sqrt{x^3}}\right)}^r$$$$\frac{6!}{r!(6-r)!}.{\left(x^{\frac{5}{2}}\right)}^{6-r}.3^r.\frac{1}{{\left(x^{\frac{3}{2}}\right)}^r}$$$$\frac{6!}{r!(6-r)!}.{\left(x\right)}^{\frac{30-5r}{2}}.3^r.\frac{1}{x^{\frac{3r}{2}}}$$$$\frac{6!\times 3^r}{r!(6-r)!}\times x^{\frac{30-5r}{2}-\frac{3r}{2}}$$$$\frac{6!\times 3^r}{r!(6-r)!}\times x^{\frac{30-8r}{2}}$$$$so{x}^{15-4r}=x^3$$$$-4r=-12sor=3$$$$\frac{6!\times 3^3}{3!3!}\times x^3$$$$requiredcoefficient=\frac{6\times 5\times 4\times 27}{6}=540$$

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