Question

please solve fst please​

Answer 1

Question: If $\sf x-\frac{1}{x}$ = 3 + 2√2, find the value of $\sf \frac{1}{4}(x^{3} - \frac{1}{x^{3}})$

Answer:

$\large{\underline{\boxed{\sf \frac{1}{4}( {x}^{3} - \frac{1}{ {x}^{3} } )=27 + 19 \sqrt{2} }}}$

Step-by-step explanation:

Given that -

$\implies \tt x - \frac{1}{x} = 3 + 2 \sqrt{2} \\ \\ \bf On \: cubing \: both \: sides : \\ \\ \implies \tt (x - \frac{1}{x} ) {}^{3} = (3 + 2 \sqrt{2} ) {}^{3} \\ \\ \implies \tt {x}^{3} - \frac{1}{ {x}^{3} } - 3(x - \frac{1}{x} ) = (3) {}^{3} + (2 \sqrt{2} ) {}^{3} + 3.3.2 \sqrt{2} (3 + 2 \sqrt{2} ) \\ \\ \implies \tt {x}^{3} - \frac{1}{ {x}^{3}} - 3(3 + 2 \sqrt{2} ) = 27 + 16 \sqrt{2} + 18 \sqrt{2} (3 + 2 \sqrt{2} ) \\ \\ \implies \tt {x}^{3} - \frac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} = 27 + 16 \sqrt{2} + 54 \sqrt{2} + 72 \\ \\ \implies \tt {x}^{3} - \frac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} = 99 + 70 \sqrt{2} \\ \\ \implies \tt {x}^{3} - \frac{1}{ {x}^{3} } = 99 + 9 + 70 \sqrt{2} + 6 \sqrt{2} \\ \\ \implies \tt {x}^{3} - \frac{1}{ {x}^{3} } = 108 + 76 \sqrt{2} \\ \\ \implies \tt {x}^{3} - \frac{1}{ {x}^{3} } = 4(27 + 19 \sqrt{2} )$

Now, we have to find $\tt \frac{1}{4}(x^{3} - \frac{1}{x^{3}})$

$\implies \tt {x}^{3} - \frac{1}{ {x}^{3} } = 4(27 + 19 \sqrt{2} ) \\ \\ \bf Multiplying \: \frac{1}{4} \: both \: sides : \\ \\ \implies \tt \frac{1}{4}( {x}^{3} - \frac{1}{ {x}^{3} } )= \frac{1}{4} \: .\: 4(27 + 19 \sqrt{2} ) \\ \\ \boxed {\bf \therefore \: \frac{1}{4}( {x}^{3} - \frac{1}{ {x}^{3} } )=27 + 19 \sqrt{2} }$

_________________________

★ Identity Used:

• (a - b)³ = a³ - b³ - 3ab (a - b)