[tex]\huge\fcolorbox{red}{red}{✪ Answer⊹❥}[/tex]
[tex]▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬[/tex]
[tex]\red\bigstar\small \tt {{Solution}}[/tex]
[tex] \huge\frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{3} [/tex]
[tex] \large⟶\frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{3} [/tex]
[tex]\large⟶ \frac{3 + \sqrt{6}( \sqrt{3} - \sqrt{2)} }{ { \sqrt{(3)} }^{2} - { \sqrt{(2)} }^{2} } = a + b \sqrt{3} [/tex]
[tex]\large⟶ \frac{ \sqrt{3} }{3 - 2} = a + b \sqrt{3} [/tex]
[tex]\large⟶ \sqrt{3} = a + b \sqrt{3} [/tex]
[tex]{\underline{\huge{\tt{\fbox{\purple{a = 0}}}}}}[/tex]
[tex]{\underline{\huge{\tt{\fbox{\green{b = 1}}}}}}[/tex]
[tex]{\underline{\huge{\tt{\fbox{\red{Thanks}}}}}}[/tex]
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Verified answer
[tex]\huge\fcolorbox{red}{red}{✪ Answer⊹❥}[/tex]
[tex]▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬[/tex]
[tex]\red\bigstar\small \tt {{Solution}}[/tex]
[tex] \huge\frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{3} [/tex]
[tex] \large⟶\frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{3} [/tex]
[tex]\large⟶ \frac{3 + \sqrt{6}( \sqrt{3} - \sqrt{2)} }{ { \sqrt{(3)} }^{2} - { \sqrt{(2)} }^{2} } = a + b \sqrt{3} [/tex]
[tex]\large⟶ \frac{ \sqrt{3} }{3 - 2} = a + b \sqrt{3} [/tex]
[tex]\large⟶ \sqrt{3} = a + b \sqrt{3} [/tex]
[tex]{\underline{\huge{\tt{\fbox{\purple{a = 0}}}}}}[/tex]
[tex]{\underline{\huge{\tt{\fbox{\green{b = 1}}}}}}[/tex]
[tex]▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬[/tex]
[tex]{\underline{\huge{\tt{\fbox{\red{Thanks}}}}}}[/tex]
I hope it will help you ✌.