Question

if xsquare +1/x sq =23,then find the value of x + 1/and xcube+ 1/xcube​

Answer 1

Step-by-step explanation:

$\large{\green{Given:}}{x}^{2} + \frac{1}{ {x}^{2} } = 23 \\ \large{\green{To \: find:}}x + \frac{1}{x} \:and\: {x}^{3} + \frac{1}{ {x}^{3 } } \\ \large{\green{Finding:}} {x}^{2} + \frac{1}{ {x}^{2} } = 23(given)\\\large{(x + \frac{1}{x} )}^{2} = {x}^{2} + 2 \times x \times \frac{1}{x} + \frac{1}{ {x}^{2} } \\\large {(x + \frac{1}{x} )}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\\large {(x + \frac{1}{x} )}^{2} = 23 + 2 = 25 \\ \large \: x + \frac{1}{x} = \sqrt{25} = 5 \\ \large{(x + \frac{1}{x}) }^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times {x}^{2} \times \frac{1}{x} + 3 \times x \times \frac{1}{ {x}^{2} } \\ \large{(x + \frac{1}{x}) }^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) \\\large {(5)}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 5 \\ \large125 = {x}^{3} + \frac{1}{ {x}^{3} } + 15 \\\large {x}^{3} + \frac{1}{ {x}^{3} } = 125 - 15 \\ \large{x}^{3} + \frac{1}{ {x}^{3} } = 110$

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