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Cube Numbers

x6-10 x3-27

Question

x⁶ $-$ 10x³ $-$ 27

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Solution

$We have x^{6} - 10 x^{3} - 27 . Here, - 10 x^{3} can not be written in cubic from, so we wil split - 10 x^{3} into - 9 x^{3} - x^{3} . x^{6} - 9 x^{3} - x^{3} - 27 = x^{6} - x^{3} - 27 - 9 x^{3} = {(x^{2})}^{3} + {(- x)}^{3} + {(- 3)}^{3} - 3 \times (x^{2}) \times (- x) \times (- 3) = [x^{2} + (- x) + (- 3)] [{(x^{2})}^{2} + {(- x)}^{2} + {(- 3)}^{2} - x^{2} \times (- x) - (- x) \times (- 3) - (- 3) \times x^{2}] = (x^{2} - x - 3) (x^{4} + x^{2} + 9 + x^{3} - 3 x + 3 x^{2}) = (x^{2} - x - 3) (x^{4} + x^{3} + 4 x^{2} - 3 x + 9)$

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Similar questions

x^{6} - 10 x^{3} - 27

f (x) = x^{6} - 10 x^{5} - 10 x^{4} - 10 x^{3} - 10 x^{2} - 10 x + 10

f (11)

x^{6} + 27

10 x^{3} - c x^{2} - 54 x - 27 = 0

H . P

c

10 x^{3} - {c x}^{2} - 54 x - 27 = 0

c

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