Question

Please help me to solve this question
Sum from algebra, class 9​

Answer 1

Step-by-step explanation:

Hello user,☺

Given that:

$\tt \red{ \bigg(x + \frac{1}{x} \bigg) ^{2} } = 3 \\ \\ </u><u>[tex] </u></p><h2><u>To </u><u>find</u><u>:</u></h2><p></p><p><u>[tex] \tt \blue{the \: valve \: of \: {x}^{3} + \frac{1}{ {x}^{3} } }$

Solution:

We have,

$\tt \red{ \bigg(x + \frac{1}{x} \bigg) ^{2} } = 3 \\ \\ </u></p><p><u>[tex] \tt \red{ \bigg(x + \frac{1}{x} \bigg) ^{2} } = 3$

$⟹ \tt \red{x + \frac{1}{x} = \sqrt{3} }$

Now,

$\tt \blue{{x}^{3} + \frac{1}{ {x}^{3} }} \\ \\ [\tex]</p><p>[tex] = \tt \blue{\bigg(x + \frac{1}{x} \bigg) ^{2} - 3x \times \frac{1}{x} \bigg (x + \frac{1}{x} \bigg)}$

$\bigg [∴ {x}^{3} + {y}^{3} = (x + y {)}^{2} - 3xy(x + y),where \: x = x,y = \frac{1}{x} \bigg ]$

$= \tt \blue{( \sqrt{3} {)}^{3} - 3 \sqrt{3} }$

$= \tt \blue{3 \sqrt{3} - 3 \sqrt{3} }$

$= \tt \blue0$

Therefore, according to the question we find the value of

$\tt \blue{{x}^{3} + \frac{1}{ {x}^{3} }} = 0 \: ans.$