Wrong questions everywhere. (Question:320113.0)

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Neil Nilesh

· started a discussion

· 1 Months ago

Wrong questions everywhere. (Question:320113.0)

There will be 3/x not 3/xcube in the first situation.

Question:

If $x^{3} + \frac{3}{x^{3}} = 4 (a^{3} + b^{3})$ and $3 x + \frac{1}{x^{3}} = 4 (a^{3} - b^{3})$ then $a^{2} - b^{2}$ is

Options:

Solution:

$\begin{matrix} 8 a^{3} = x^{3} + \frac{3}{x^{3}} + 3 x + \frac{1}{x^{3}} \\ = x^{3} + 3 (x^{2} \times \frac{1}{x}) + 3 (x \times \frac{1}{x^{2}}) + \frac{1}{x^{3}} \\ = 3 {(x + \frac{1}{x})}^{3} \\ 2 a = x + \frac{1}{x} . . . . (i) \\ I n t h e s a m e w a y \\ 2 b = x - \frac{1}{x} . . . . . (i i) \\ F r o m e q . (i) a n d (i i) \\ 4 (a^{2} + b^{2}) = 4 \\ (a^{2} - b^{2}) = 1 \end{matrix}$

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Wrong questions everywhere. (Question:320113.0)

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