Factorize $a^{6} - 7 a^{3} - 8$

(a - 4) (a^{2} + 2 a + 4) (a + 2) (a^{2} - a + 1)

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(a - 2) (a^{2} + 2 a + 4) (a + 1) (a^{2} - a + 1)

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(a - 2) (a^{2} + a + 2) (a - 1) (a^{2} - a + 1)

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(a + 2) (a^{2} + 2 a + 4) (a + 1) (a^{2} - a + 1)

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Solution

The correct option is B $(a - 2) (a^{2} + 2 a + 4) (a + 1) (a^{2} - a + 1)$

$a^{6} - 7 a^{3} - 8 = (a^{3})^{2} - 7 a^{3} - 8 = (a^{3})^{2} - 8 a^{3} + a^{3} - 8 = (a^{3} - 8) (a^{3} + 1) = (a - 2) (a^{2} + 2 a + 4) (a + 1) (a^{2} - a + 1)$

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Factorise $2 a^{7} - 128 a$

{(a + \frac{1}{2 a +} \frac{1}{2 a +} \dots)}^{2} + {(a - \frac{1}{2 a -} \frac{1}{2 a -} \dots)}^{2} = 2 a^{2}

(a + \frac{1}{2 a +} \frac{1}{2 a +} \dots) (a - \frac{1}{2 a -} \frac{1}{2 a -} \dots) = a^{2} - \frac{1}{2 a^{2} -} \frac{1}{2 a^{2} -} \dots

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a^{2} b (a^{3} - a + 1) - a b (a^{4} - 2 a^{2} + 2 a) - b (a^{3} - a^{2} - 1)

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