What is (x+2)/(x^2+x+1)-(1-x^2)/(x^3-1) ?

The solution to (x+2)/(x^2+x+1)-(1-x^2)/(x^3-1) is (2x+3)/(x^2+x+1)

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simplify (x+2)/(x^2+x+1)-(1-x^2)/(x^3-1)

simplify

\frac{x + 2}{x ^{2} + x + 1} - \frac{1 - x ^{2}}{x ^{3} - 1}

\frac{2 x + 3}{x ^{2} + x + 1}

Solution steps

\frac{x + 2}{x ^{2} + x + 1} - \frac{1 - x ^{2}}{x ^{3} - 1}

Cancel

\frac{1 - x ^{2}}{x ^{3} - 1} : - \frac{1 + x}{x ^{2} + x + 1}

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

Apply rule:

- (- a) = a

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

= \frac{x + 2}{x ^{2} + x + 1} + \frac{1 + x}{x ^{2} + x + 1}

Apply the fraction rule:

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

= \frac{x + 2 + 1 + x}{x ^{2} + x + 1}

Simplify

x + 2 + 1 + x : 2 x + 3

= \frac{2 x + 3}{x ^{2} + x + 1}

f (x + 2)

\frac{- x ^{2}}{x}

- 1 (- 1)

i^{78}

\frac{4 c}{6} + \frac{2 c}{6}

What is (x+2)/(x^2+x+1)-(1-x^2)/(x^3-1) ?
The solution to (x+2)/(x^2+x+1)-(1-x^2)/(x^3-1) is (2x+3)/(x^2+x+1)