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Multiplying the fraction by $x+1$
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$\frac{\frac{\left(x^3-1\right)\left(x+1\right)}{x^3-2x^2-3x}}{x^2+x-2}+\frac{x^2+x+1}{6x+x^2-x^3}$
Learn how to solve problems step by step online. Simplify ((x^3-1)/(x^3-2x^2-3x)(x+1))/(x^2+x+-2)+(x^2+x+1)/(6x+x^2-x^3). Multiplying the fraction by x+1. Divide fractions \frac{\frac{\left(x^3-1\right)\left(x+1\right)}{x^3-2x^2-3x}}{x^2+x-2} with Keep, Change, Flip: \frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}. We can factor the polynomial \left(x^3-2x^2-3x\right) using the rational root theorem, which guarantees that for a polynomial of the form a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 there is a rational root of the form \pm\frac{p}{q}, where p belongs to the divisors of the constant term a_0, and q belongs to the divisors of the leading coefficient a_n. List all divisors p of the constant term a_0, which equals 0. Next, list all divisors of the leading coefficient a_n, which equals 1.