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प्रश्न
Using binomial theorem, write down the expansions :
(vi) \[\left( \frac{\sqrt{x}}{a} - \sqrt{\frac{a}{x}} \right)^6\]
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उत्तर
(vi) \[\left( \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}} \right)^6 \]
\[ = ^{6}{}{C}_0 \left( \sqrt{\frac{x}{a}} \right)^6 \left( \sqrt{\frac{a}{x}} \right)^0 - ^{6}{}{C}_1 \left( \sqrt{\frac{x}{a}} \right)^5 \left( \sqrt{\frac{a}{x}} \right)^1 + ^{6}{}{C}_2 \left( \sqrt{\frac{x}{a}} \right)^4 \left( \sqrt{\frac{a}{x}} \right)^2 - ^{6}{}{C}_3 \left( \sqrt{\frac{x}{a}} \right)^3 \left( \sqrt{\frac{a}{x}} \right)^3 +^{6}{}{C}_4 \left( \sqrt{\frac{x}{a}} \right)^2 \left( \sqrt{\frac{a}{x}} \right)^4 -^{6}{}{C}_5 \left( \sqrt{\frac{x}{a}} \right)^1 \left( \sqrt{\frac{a}{x}} \right)^5 + ^{6}{}{C}_6 \left( \sqrt{\frac{x}{a}} \right)^0 \left( \sqrt{\frac{a}{x}} \right)^6 \]
\[ = \frac{x^3}{a^3} - 6\frac{x^2}{a^2} + 15\frac{x}{a} - 20 + 15\frac{a}{x} - 6\frac{a^2}{x^2} + \frac{a^3}{x^3}\]
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