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Answer 1

Answer:

$\large{\underline{\boxed{\sf {a}^{3} + {b}^{3} = 970}}}$

Step-by-step explanation:

$\sf a = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \: and \: \: b = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

Here, we will rationalise the denominator of a first, and then b.

$\implies \sf a = \frac{ \sqrt{3} + \sqrt{2}} { \sqrt{3} - \sqrt{2} } \\ \\ \implies \sf \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2}}{ \sqrt{3 + \sqrt{2} } } \\ \\ \implies \sf \frac{( \sqrt{3} + \sqrt{2}) {}^{2} }{( \sqrt{3}) {}^{2} - {( \sqrt{2} )}^{2} } \\ \\ \implies \sf \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ \implies \sf a = 5 + 2 \sqrt{6}$

Now, rationalise the denominator of b:

$\sf b = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \implies \sf \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \implies \sf \frac{( \sqrt{3} - \sqrt{2} ) {}^{2} }{( \sqrt{3} ) {}^{2} - {( \sqrt{2} )}^{2} } \\ \\ \implies \sf \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ \implies \sf b = 5 - 2 \sqrt{6}$

Find the product of a and b.

$\implies \sf ab = (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ \implies \sf ab = (5) {}^{2} - (2 \sqrt{6} ) {}^{2} \\ \\ \implies \sf ab = 25 - 24 \\ \\ \implies \sf ab = 1$

Now, find (a + b):

$\implies \sf a + b = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} \\ \\ \implies \sf a + b = 5 + 5 \\ \\ \implies \sf a + b = 10 \\ \\ \bf on \: cubing \: both \: sides - \\ \\ \implies \sf (a + b) {}^{3} = (10) {}^{3} \\ \\ \implies \sf {a}^{3} + {b}^{3} + 3ab(a + b) = 1000 \\ \\ \implies \sf {a}^{3} + {b}^{3} + 3 \times 1(10) = 1000 \\ \\ \implies \sf {a}^{3} + {b}^{3} + 30 = 1000 \\ \\ \implies \sf {a}^{3} + {b}^{3} = 1000 - 30 \\ \\ \boxed{ \therefore \bf {a}^{3} + {b}^{3} = 970}$

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\bold{{a}^{3} + {b}^{3} = 970}}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :

$a = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: and \: b = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

To find : a³ + b³

Solution :

First rationalize the denominators of a and b

Consider a

The rationalising factor of √3 - √2 is √3 + √2. So, multiply numerator and denominator with rationalising factor.

$\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$= \dfrac{ {( \sqrt{3} + \sqrt{2}) }^{2} }{ {( \sqrt{3})}^{2} - {( \sqrt{2})}^{2} }$

Since (x + y)(x - y) = x² - y²

$= \dfrac{ {( \sqrt{3})}^{2} + 2( \sqrt{3})( \sqrt{2}) + {( \sqrt{2})}^{2} }{3 - 2}$

Since (x + y)² = x² + 2xy + y²

$= \dfrac{3 + 2 \sqrt{6} + 2}{1}$

$= 5 + 2 \sqrt{6}$

$\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = 5 + 2\sqrt{6}$

Consider b

The rationalising factor of √3 + √2 is √3 - √2. So, multiply numerator and denominator with rationalising factor.

$\dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$= \dfrac{ {( \sqrt{3} - \sqrt{2} )}^{2}}{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} }$

Since (x + y)(x - y) = x² - y²

$= \dfrac{ {( \sqrt{3}) }^{2} - 2( \sqrt{3})( \sqrt{2}) + {( \sqrt{2}) }^{2} }{3 - 1}$

Since (x - y)² = x² - 2xy + y²

$= \dfrac{3 - 2 \sqrt{6} + 2}{1}$

$= 5 - 2 \sqrt{6}$

$\dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = 5 - 2\sqrt{6}$

Find the value of (ab)

$ab = (5 + 2 \sqrt{6})(5 - 2 \sqrt{6})$

Since (x + y)(x - y) = x² - y²

$ab = {5}^{2} - {(2 \sqrt{6})}^{2}$

$ab = 25 - 4(6)$

$ab = 25 - 24$

$ab = 1$

Find the value of a + b

$a + b = 5 + 2\sqrt{6} + (5 - 2 \sqrt{6})$

$a + b = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6}$

$a + b = 5 + 5$

$a + b = 10$

By cubing on both sides

${(a + b)}^{3} = {10}^{3}$

${a}^{3} + {b}^{3} + 3ab(a + b) = 1000$

Since (x + y)³ = x³ + y³ + 3xy(x + y)

By substituting the value of a + b and ab:

${a}^{3} + {b}^{3} + 3 \times 1(10) = 1000$

${a}^{3} + {b}^{3} + 3(10) = 1000$

${a}^{3} + {b}^{3} + 30 = 1000$

${a}^{3} + {b}^{3} = 1000 - 30$

${a}^{3} + {b}^{3} = 970$

$\boxed{\bold{{a}^{3} + {b}^{3} = 970}}$

Identities used :-

(x + y)(x - y) = x² - y²

(x + y)² = x² + 2xy + y²

(x - y)² = x² - 2xy + y²

(x + y)³ = x³ + y³ + 3xy(x + y)

Extra info related to the question :-

What is a rationlising factor ?

If the product of two irrational numbers is a rational number then each of the the two is rationalising factor of the other.