If $a + b = 1$ and $a^{2} + b^{2} = 2$ , then find the value of $a^{8} + b^{8}$ .

\frac{97}{8}

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\frac{87}{8}

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\frac{97}{8}

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\frac{87}{9}

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Solution

The correct option is A $\frac{97}{8}$
Given, $a + b = 1$ and $a^{2} + b^{2} = 2$

W.K.T, $(a + b)^{2} = a^{2} + b^{2} + 2 a b$
$\Rightarrow (1)^{2} = 2 + 2 a b$
$\Rightarrow a b = \frac{- 1}{2}$

$(a^{2} + b^{2})^{2} = a^{4} + b^{4} + 2 a^{2} b^{2}$
$\Rightarrow a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2 (a b)^{2}$
$\Rightarrow a^{4} + b^{4} = (2)^{2} - 2 (\frac{- 1}{2})^{2}$
$\Rightarrow a^{4} + b^{4} = 4 - \frac{1}{2}$
$\Rightarrow a^{4} + b^{4} = \frac{7}{2}$

$(a^{4} + b^{4})^{2} = a^{8} + b^{8} + 2 a^{4} b^{4}$
$\Rightarrow a^{8} + b^{8} = (a^{4} + b^{4})^{2} - 2 (a b)^{4}$
$\Rightarrow a^{8} + b^{8} = (\frac{7}{2})^{2} - 2 (\frac{- 1}{2})^{4}$
$\Rightarrow a^{8} + b^{8} = \frac{49}{4} - \frac{1}{8}$
$\Rightarrow a^{8} + b^{8} = \frac{98 - 1}{8}$
$\Rightarrow a^{8} + b^{8} = \frac{97}{8}$

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If a+b=1 and a2+b2=2, then find the value of a8+b8.

If $a + b = 1$ and $a^{2} + b^{2} = 2$ , then find the value of $a^{8} + b^{8}$ .