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

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

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

simplify

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

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

Solution steps

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

Cancel

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

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

Apply rule:

- (- a) = a

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

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

Apply the fraction rule:

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

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

Simplify

m + 2 + 1 + m : 2 m + 3

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

lo g_{2} (\frac{3}{4})

(2 - 2 i)^{3}

\frac{x ^{2} - 6 x - 16}{3 x ^{3} + 6 x ^{2}}

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

\frac{4}{3} + \frac{5}{2}

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