Class : 9 ( Ninth )Topic : Rationalisation of denominator Solve with p

1.	Class : 9 ( Ninth )Topic : Rationalisation of denominator Solve with process ❤️Please Help ☹️✅
Answer» $\large\underline{\sf{Solution-}}$ Given that, $\red{\rm :\longmapsto\:x = \dfrac{1}{7 + 4 \sqrt{3} }}$ Multiply and Divide by conjugate of denominator, we get $\: \: \rm = \: \: \dfrac{1}{7 + 4 \sqrt{3} } \times \dfrac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$ $\: \: \rm = \: \: \dfrac{7 - 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} }$ $\: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} }$ $\: \: \rm = \: \: \dfrac{7 - 4 \sqrt{3} }{49 - 48}$ $\: \: \rm = \: \: 7 - 4 \sqrt{3}$ $\: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:x = 7 - 4 \sqrt{3}}}$ Also, Given that $\green{\rm :\longmapsto\:y = \dfrac{1}{7 - 4 \sqrt{3} }}$ Multiply and Divide by conjugate of denominator, we get $\: \: \rm = \: \: \dfrac{1}{7 - 4 \sqrt{3} } \times \dfrac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} }$ $\: \: \rm = \: \: \dfrac{7 + 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} }$ $\: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} }$ $\: \: \rm = \: \: \dfrac{7 + 4 \sqrt{3} }{49 - 48}$ $\: \: \rm = \: \: 7 + 4 \sqrt{3}$ $\: \: \: \: \: \: \: \green{ \boxed{\bf\implies \:y = 7 + 4 \sqrt{3}}}$ CONSIDER, $\red{\rm :\longmapsto\:xy}$ $\: \: \rm = \: \: (7 + 4 \sqrt{3})(7 - 4 \sqrt{3})$ $\: \: \rm = \: \: {(7)}^{2} - {(4 \sqrt{3}) }^{2}$ $\: \: \rm = \: \: 49 - 48$ $\: \: \rm = \: \: 1$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:xy = 1}}$ Consider, $\blue{\bf :\longmapsto\:x + y}$ $\: \: \rm = \: \: 7 + 4 \sqrt{3} + 7 - 4 \sqrt{3}$ $\: \: \rm = \: \: 14$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \blue{ \boxed{\bf\implies \:x + y = 14}}$ Now, we have to find the value of $\green{\bf :\longmapsto\: {x}^{3} + {y}^{3} }$ $\: \: \rm = \: \: {(x + y)}^{3} - 3xy(x + y)$ $\: \: \rm = \: \: {14}^{3} - 3 \times 1 \times 14$ [ using above VALUES of x + y = 14 and xy = 1 ] $\: \: \rm = \: \: 2744 - 42$ $\: \: \rm = \: \: 2702$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{ \boxed{\bf\implies \: {x}^{3} + {y}^{3} = 2704}}$ Additional Information :- More IDENTITIES to know: (a + b)² = a² + 2AB + b² (a - b)² = a² - 2ab + b² a² - b² = (a + b)(a - b) (a + b)² = (a - b)² + 4ab (a - b)² = (a + b)² - 4ab (a + b)² + (a - b)² = 2(a² + b²) (a + b)³ = a³ + b³ + 3ab(a + b) (a - b)³ = a³ - b³ - 3ab(a - b)