If a $^{2}$ +1 = a, then the value of a $^{8}$ +a $^{6}$ +1 is, if a = (2) $^{1 / 7}$

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Solution

The correct option is A
3

$a^{7} = 2, a^{2} + 1 = a$

$a^{8} + a^{6} + 1 = a^{6} (a^{2} + 1) + 1 = \frac{2}{a} (a) + 1 = 2 + 1 = 3$

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Similar questions

{(1 + x)}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . a_{10} x^{10}

{(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})}^{2} + {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2}

(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

a_{1}, a_{2}, a_{3}, \dots \dots

a_{3} + a_{5} + a_{8} = 11

a_{4} + a_{2} = - 2

a_{1} + a_{6} + a_{7}

\frac{a^{2} - 1}{a} = 2

\frac{a^{6} - 1}{a^{3}}

I f (1 + x)^{10} = a_{0} + a_{1} x + . . . . a_{10} x^{10}, t h e n (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

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