Question

$if \: a \: = \sqrt{6} - \sqrt{3} \: and \: b \: = \: \sqrt{3} - \sqrt{2} \: and \: c \: = \: \sqrt{2} - \sqrt{6} \: then \: find \: the \: value \: of \: {a}^{3} + \: {b}^{3} + {c}^{3} - 2abc \: .$
$plz \: answer \: with \: correct \: explanatiom$

​

Answer 1

Step-by-step explanation:

\sf{\frac{ \sqrt{5 } - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } = a + b \sqrt{15} }

5

+

3

5

−

3

=a+b

15

rationalise the denominator

\sf\frac{ \sqrt{5 } - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a + b \sqrt{15}

5

+

3

5

−

3

×

5

−

3

5

−

3

=a+b

15

\sf\frac{{ (\sqrt{5 } - \sqrt{3})}^{2} }{{( \sqrt{5})}^{2} - {(\sqrt{3}) }^{2}} = a + b \sqrt{15}

(

5

)

2

−(

3

)

2

(

5

−

3

)

2

=a+b

15

\sf\frac{{ {(\sqrt{5 } )}^{2} + (\sqrt{3})}^{2} - 2( \sqrt{5} )( \sqrt{3} )}{{5} - 3} = a + b \sqrt{15}

5−3

(

5

)

2

+(

3

)

2

−2(

5

)(

3

)

=a+b

15

\sf\frac{5 + 3 - 2\sqrt{15}}{5 - 3} = a + b \sqrt{15}

5−3

5+3−2

15

=a+b

15

\sf\frac{8 - 2\sqrt{15}}{2} = a + b \sqrt{15}

2

8−2

15

=a+b

15

\sf{ \frac{2(4 + \sqrt{15} )}{2} = a + b \sqrt{15} }

2

2(4+

15

)

=a+b

15

\sf{4 + \sqrt{15} = a + b \sqrt{15} }4+

15

=a+b

15

on comparing both sides

\begin{gathered} \sf{a = 4} \\ \sf{b = 1}\end{gathered}

a=4

b=1

Answer 2

Explanation:

$a {}^{3} + b {}^{3} + c {}^{3} - 2abc$

$= a { }^{3} + b {}^{3} + c {}^{3} - 3abc + abc$

$= (a + b + c)(a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) + abc$

$= ( \sqrt{6} - \sqrt{3} + \sqrt{3} - \sqrt{2} + \sqrt{2} - \sqrt{6} )( {a}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) + abc$

$= 0( {a}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) + abc$

$= 0 + abc$

$= abc$

$= ( \sqrt{6} - \sqrt{3} )( \sqrt{3} - \sqrt{2} )( \sqrt{2} - \sqrt{6} )$