What is factor 2(x-y)^3+x^{-3}-y^3 ?

The solution to factor 2(x-y)^3+x^{-3}-y^3 is (1+2x^6-6x^5y+6x^4y^2-3x^3y^3)/(x^3)

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factor 2(x-y)^3+x^{-3}-y^3

factor

2 (x - y)^{3} + x^{- 3} - y^{3}

\frac{1 + 2 x ^{6} - 6 x ^{5} y + 6 x ^{4} y ^{2} - 3 x ^{3} y ^{3}}{x ^{3}}

Solution steps

2 (x - y)^{3} + x^{- 3} - y^{3}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Convert element to fraction:

2 (x - y)^{3} = \frac{2 ( x - y ) ^{3} x ^{3}}{x ^{3}}, y^{3} = \frac{y ^{3} x ^{3}}{x ^{3}}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Expand

1 + 2 (x - y)^{3} x^{3} - y^{3} x^{3} : 1 + 2 x^{6} - 6 x^{5} y + 6 x^{4} y^{2} - 3 y^{3} x^{3}

= \frac{1 + 2 x ^{6} - 6 x ^{5} y + 6 x ^{4} y ^{2} - 3 x ^{3} y ^{3}}{x ^{3}}

(x + 1) + (2 x^{2})

x^{2} - 9 x + 20.25

(x + 9)^{3} - y^{3}

[(5 x + 3) + 2 y] [(5 x + 3) - 2 y]

cos^{2} (y) + 2 cos (y) + 1

What is factor 2(x-y)^3+x^{-3}-y^3 ?
The solution to factor 2(x-y)^3+x^{-3}-y^3 is (1+2x^6-6x^5y+6x^4y^2-3x^3y^3)/(x^3)