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The sum of a number and its inverse is −4. The sum of their cubes is :

Let \( x + \frac{1}{x} = -4 \). Using \( a^3 + b^3 = (a + b)^3 - 3ab(a + b) \) with \( ab = x \cdot \frac{1}{x} = 1 \), we get \( x^3 + \frac{1}{x^3} = (-4)^3 - 3(-4) = -64…

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The sum of a number and its inverse is −4. The sum of their cubes is :

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Correct answer A

Let x+1x=−4 x + \frac{1}{x} = -4 . Using a3+b3=(a+b)3−3ab(a+b) a^3 + b^3 = (a + b)^3 - 3ab(a + b) with ab=x⋅1x=1 ab = x \cdot \frac{1}{x} = 1 , we get x3+1x3=(−4)3−3(−4)=−64+12=−52 x^3 + \frac{1}{x^3} = (-4)^3 - 3(-4) = -64 + 12 = -52 . The value −64 forgets the −3ab(a+b) -3ab(a + b) term, and 52 has the wrong sign.

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