Question

If $x + \frac{1}{x} = 7$<br />then find the value of ${x}^{3} + \frac{1}{x {}^{3} }$​. No spam please​

Answer 1

Step-by-step explanation:

चाणक्य ने चंद्रगुप्त की मां से क्या कहा

Answer 2

Given:

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$x + \frac{1}{x} = 7$

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To find:

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${x}^{3} + \frac{1}{x {}^{3} }$

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Solution:

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Using identity of $(a + b) {}^{3}$

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$( x + \frac{1}{x} ) {}^{3} = x {}^{3} + \frac{1}{x {}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} )$

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$\implies \: (x + \frac{1}{x} ) {}^{3} = x {}^{3} + \frac{1}{x {}^{3} } + 3(x + \frac{1}{x} )$

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Now keeping the value of x + 1/x = 7, we get

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$7 {}^{3} = x {}^{3} + \frac{1}{x {}^{3} } + 3 \times 7 \\ \\ \implies343 = {x}^{3} + \frac{1}{x {}^{3} } + 21 \\ \\ \implies \: {x}^{3} + \frac{1}{x} {}^{3} = 343 - 21 \\ \\ \implies \: x {}^{3} + \frac{1}{x {}^{3} } = 322$

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Hence, 322 is the required answer.