Question

Please try to solve....​

Answer 1

Required Answer:-

Given:

$\rm \mapsto x + \frac{1}{x} + 2 = 0$

To find:

Value of $\rm {x}^{33} + {x}^{32} + {x}^{13} + {x}^{12} + x + 1$

Solution:

Given that,

$\rm \implies x + \dfrac{1}{x} + 2 =0$

$\rm \implies x + \dfrac{1}{x} = - 2$

$\rm \implies \dfrac{ {x}^{2} + 1}{x} = - 2$

$\rm \implies {x}^{2} + 1 = - 2x$

$\rm \implies {x}^{2} + 2x + 1 = 0$

$\rm \implies {x}^{2} +x + x + 1 = 0$

$\rm \implies x(x + 1) + 1(x+ 1) = 0$

$\rm \implies (x + 1)(x+ 1) = 0$

$\rm \implies (x + 1)^{2} = 0$

$\rm \implies (x + 1) = 0$

$\rm \implies x = - 1$

Hence, we have found the value of x.

Therefore,

$\rm{x}^{33} + {x}^{32} + {x}^{13} + {x}^{12} + x + 1$

$\rm = {( - 1)}^{33} + {( - 1)}^{32} + {( - 1)}^{13} + {( - 1)}^{12} - 1 + 1$

$\rm = - 1 + 1 - 1 + 1 - 1 + 1$

$\rm = 3 - 3$

$\rm = 0$

Hence, the required answer is 0.

So,

$\rm \mapsto {x}^{33} + {x}^{32} + {x}^{13} + {x}^{12} + x + 1 = 0$

Answer 2

Answer:

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