Find the values of a and b if √7+1/√7-1 - √7-1/√7+1 =a+b√7.​

Answer:

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\bf \: a=0\qquad \: \\ \\& \qquad \:\bf \: b= \frac{2}{3} \end{aligned}} \qquad \\ \\ [/tex]

Step-by-step explanation:

Given expression is

[tex]\sf \:\dfrac{ \sqrt{7} + 1}{ \sqrt{7} - 1} - \dfrac{ \sqrt{7} - 1}{ \sqrt{7} + 1} = a + b \sqrt{7} \\ \\ [/tex]

On taking LCM, we get

[tex]\sf \:\dfrac{( \sqrt{7} + 1)^{2} - {( \sqrt{7} - 1)}^{2} }{ (\sqrt{7} - 1)( \sqrt{7} + 1)} = a + b \sqrt{7} \\ \\ [/tex]

We know,

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \qquad \: \\ \\& \qquad \:\sf \: (x + y)(x - y)= {x}^{2} - {y}^{2} \end{aligned}} \qquad \\ \\ [/tex]

So, using these algebraic identities, we get

[tex]\sf \:\dfrac{4( \sqrt{7})(1)}{ {( \sqrt{7}) }^{2} - {(1)}^{2} } = a + b \sqrt{7} \\ \\ [/tex]

[tex]\sf \:\dfrac{4 \sqrt{7}}{7 - 1} = a + b \sqrt{7} \\ \\ [/tex]

[tex]\sf \:\dfrac{4 \sqrt{7}}{6} = a + b \sqrt{7} \\ \\ [/tex]

[tex]\sf \:\dfrac{2\sqrt{7}}{3} = a + b \sqrt{7} \\ \\ [/tex]

On comparing, we get

[tex]\bf\implies \: a = 0 \: \: \: and \: \: \: b = \dfrac{2}{3} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]