Question

The sum of their number and its inverse is -4. the sum of their cubes

Answer 1

Step-by-step explanation:

Suppose that the number is x and its inverse is 1/x.

Then, according to the statement of the problem

$x + \frac{1}{x} = - 4$

On cubing both sides we get

${(x + \frac{1}{x}) }^{3} = - 64$

which on expansion gives

${x}^{3} + \frac{1}{ {x}^{3} } + 3 {x}^{2} \times \frac{1}{x} + 3x \times \frac{1}{ {x}^{2} } = - 64 \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = - 64 \\ {x}^{3} + \frac{1}{ {x}^{3} } - 12 = - 64 \\ {x}^{3} + \frac{1}{ {x}^{3} } = - 48$

which is the required value.

Answer 2

The sum of their cubes is -52

Solution:

Let "x" be the number

To find: $x^3 + \frac{1}{x^3}$

The sum of their number and its inverse is -4

$Inverse = \frac{1}{x}$

Therefore,

$x + \frac{1}{x} = -4$

Take cube on both sides

$(x + \frac{1}{x})^3 = (-4)^3\\\\(x + \frac{1}{x})^3 = - 64$

Use the following identity,

$(a+b)^3 = a^3+b^3+3ab(a+b)$

Therefore,

$x^3 + (\frac{1}{x})^3 + 3 \times x \times \frac{1}{x} \times (x + \frac{1}{x}) = -64\\\\x^3 + \frac{1}{x^3} + 3 \times (x + \frac{1}{x}) = -64\\\\Substitute\ (x + \frac{1}{x}) = -4\\\\x^3 + \frac{1}{x^3} + 3 \times -4 = -64\\\\x^3 + \frac{1}{x^3} -12 = -64\\\\x^3 + \frac{1}{x^3} = -64 + 12\\\\x^3 + \frac{1}{x^3} = -52$

Thus sum of their cubes is -52

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