Question

$if \: \:a+\frac{1}{a}=\sqrt{3} \: \: then \: what \: is \: the \: value \: of \: {a}^{30} + {a}^{24} + {a}^{18} + {a}^{12} + {a}^{6} + 1 \:$
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Answer 1

$a + \dfrac{1}{a} = \sqrt{3}$

Squaring Both Sides,

${a}^{2} + \dfrac{1}{ {a}^{2} } + 2= 3$

${a}^{2} + \dfrac{1}{ {a}^{2} } = 1$

Cubing Both Sides,

${a}^{6} + \dfrac{1}{ {a}^{6} } + 3 \times {a}^{2} \times \dfrac{1}{ {a}^{2} } ( {a}^{2} + \dfrac{1}{ {a}^{2} } ) = 1$

${a}^{6} + \dfrac{1}{ {a}^{6} } = - 2$

Hence,

${a}^{6} = - 1$

To Find:

${a}^{30} + {a}^{24} + {a}^{18} + {a}^{12} + {a}^{6} + 1 \\ = ( {a}^{6} ) ^{5} + ({a}^{6} ) ^{4} + ( {a}^{6} ) ^{3} +( {a}^{6} ) ^{2} + ( {a}^{6} ) ^{1} + ( {a}^{6} ) ^{0} \\ = {( - 1)}^{5} + ( - 1) ^{4} + ( - 1) ^{3} + ( - 1) ^{2} + ( - 1) ^{1} + ( - 1) ^{0} \\ = - 1 + 1 - 1 + 1 - 1 + 1 = 0$

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