Rationalize the denominator:
1(6−3)\dfrac{1}{(\sqrt{6} - \sqrt{3})}(6−3)1
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Rationalizing the denominator,
⇒1(6−3)×(6+3)(6+3)⇒(6+3)(6)2−(3)2⇒(6+3)6−3⇒(6+3)3\Rightarrow \dfrac{1}{(\sqrt{6} - \sqrt{3})} \times \dfrac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{6 - 3} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{3} \\[1em]⇒(6−3)1×(6+3)(6+3)⇒(6)2−(3)2(6+3)⇒6−3(6+3)⇒3(6+3)
Hence, on rationalizing 1(6−3)=(6+3)3\dfrac{1}{(\sqrt{6} - \sqrt{3})} = \dfrac{(\sqrt{6} + \sqrt{3})}{3}(6−3)1=3(6+3).
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