Question

$if \: {x}^{2} + \frac{1}{ {x}^{2} } = 7 \:find \: the \: value \: of \: 7 {x}^{3} + 8x - \frac{7}{ {x}^{3} } - \frac{8}{x}$
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Answer 1

$\huge{\mathfrak{Answer:}}$

Let x² + (1/x²) = 7 ...... (1) 

Then, 

∴ ( x² + 1/x² - 2.x.1/x ) = 7 - 2 

∴ ( x - 1/x )² = 5 

∴ x - 1/x = ± √5 .............. (2) 

∴ x³ - 1/x³ = ( x - 1/x )( x² + 1/x² + 1 ) 
= ( ± √5 ) [ ( 7 ) + 1 ] ............. from (1), (2) 

. . . . . . . . .= ± 8√5 .................. (3) 

Hence, the required expression is 

➡ 7x³ + 8x - 7/x³ - 8x 
➡ 7 ( x³ - 1/x³ ) + 8 ( x - 1/x ) 
➡ 7 ( ± 8√5 ) + 8 ( ± √5 ) 
➡ ± ( 56 + 8 ) √5 
➡ ± 64√5 

Answer 2

Solution :

Given : $\displaystyle \mathsf{x^{2}+\frac{1}{x^{2}}=7}$

$\displaystyle \implies \mathsf{(x-\frac{1}{x})^{2}+2.x.\frac{1}{x}=7}$

$\displaystyle \implies \mathsf{(x-\frac{1}{x})^{2}=5=(\sqrt{5})^{2}}$

$\displaystyle \implies \mathsf{x-\frac{1}{x}=\pm \sqrt{5}}$

Now, $\displaystyle \mathsf{7x^{3}+8x-\frac{7}{x^{3}}-\frac{8}{x}}$

$\displaystyle \mathsf{=7(x^{3}-\frac{1}{x^{3}})+8(x-\frac{1}{x})}$

$\displaystyle \mathsf{=7\bigg[(x-\frac{1}{x})^{3}+3.x.\frac{1}{x}(x-\frac{1}{x})\bigg]+8(x-\frac{1}{x})}$

$\displaystyle \mathsf{=7\bigg[(x-\frac{1}{x})^{3}+3(x-\frac{1}{x})\bigg]+8(x-\frac{1}{x})}$

$\displaystyle \mathsf{=7\{(\pm \sqrt{5})^{3}+3(\pm \sqrt{5})\}+8(\pm \sqrt{5})}$

$\displaystyle \mathsf{=\pm 7(5\sqrt{5}+3\sqrt{5})\pm 8\sqrt{5}}$

$\displaystyle \mathsf{=\pm \{7(8\sqrt{5})+8\sqrt{5}\}}$

$\displaystyle \mathsf{=\pm (56\sqrt{5}+8\sqrt{5})}$

$\displaystyle \mathsf{=\pm 64\sqrt{5}}$

$\displaystyle \to \boxed{\mathsf{7x^{3}+8x-\frac{7}{x^{3}}-\frac{8}{x}=\pm 64\sqrt{5}}}$