How do you expand ${(x^{3} - \frac{1}{x^{2}})}^{3}$ $(x^3-1/x^2)^3$ ?

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How do you expand ${(x^{3} - \frac{1}{x^{2}})}^{3}$ $(x^3-1/x^2)^3$ ?

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Answer 1 · 2018-06-12T12:19:40

$x^{9} - 3 x^{4} + \frac{3}{x} - \frac{1}{x^{6}}$ $x^9-3x^4+3/x-1/x^6$

Explanation:

now

${(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - - (1)$ $(a+b)^3=a^3+3a^2b+3ab^2+b^3--(1)$

${(x^{3} - \frac{1}{x^{2}})}^{3}$ $(x^3-1/x^2)^3$

$a = x^{3}, b = - \frac{1}{x^{2}}$ $a=x^3, b=-1/x^2$

substituting into $(1)$ $(1)$

$= {(x^{3})}^{3} + 3 {(x^{3})}^{2} (- \frac{1}{x^{2}}) + 3 x^{3} {(- \frac{1}{x^{2}})}^{2} + {(- \frac{1}{x^{2}})}^{3}$ $=(x^3)^3+3(x^3)^2(-1/x^2)+3x^3(-1/x^2)^2+(-1/x^2)^3$

now simplify

$= x^{9} - 3 \frac{x^{6}}{x^{2}} + 3 \frac{x^{3}}{x^{4}} - \frac{1}{x^{6}}$ $=x^9-3x^6/x^2+3x^3/x^4-1/x^6$

$= x^{9} - 3 x^{4} + \frac{3}{x} - \frac{1}{x^{6}}$ $=x^9-3x^4+3/x-1/x^6$

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How do you expand ${(x^{3} - \frac{1}{x^{2}})}^{3}$ $(x^3-1/x^2)^3$ ?

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