1. **State the problem:** We want to find constants $a$ and $b$ such that $$\frac{1 - x + 6x^2}{x - x^3} = \frac{a}{x - 2} - \frac{b}{x - 1}.$$ 2. **Rewrite the denominator:** Note that $$x - x^3 = x(1 - x^2) = x(1 - x)(1 + x),$$ but since the right side has denominators $x-2$ and $x-1$, we will clear denominators by multiplying both sides by $(x-2)(x-1)(x - x^3)$ to find $a$ and $b$. 3. **Multiply both sides by the common denominator:** Multiply both sides by $(x - 2)(x - 1)(x - x^3)$: $$ (1 - x + 6x^2)(x - 2)(x - 1) = a(x - x^3)(x - 1) - b(x - x^3)(x - 2).$$ 4. **Simplify the right side:** Recall $x - x^3 = x(1 - x^2) = x(1 - x)(1 + x)$. 5. **Evaluate at convenient values to solve for $a$ and $b$: ** - Let $x=2$ (to eliminate $b$ term): $$\frac{1 - 2 + 6(2)^2}{2 - 2^3} = \frac{a}{2 - 2} - \frac{b}{2 - 1}$$ The left side denominator is $2 - 8 = -6$, numerator is $1 - 2 + 24 = 23$, so left side is $\frac{23}{-6} = -\frac{23}{6}$. Right side at $x=2$ is $\frac{a}{0} - \frac{b}{1}$, undefined for $a$ term, so better to use the equation after clearing denominators. Using the cleared equation: At $x=2$: $$(1 - 2 + 6(2)^2)(2 - 2)(2 - 1) = a(2 - 2^3)(2 - 1) - b(2 - 2^3)(2 - 2)$$ Left side: $(1 - 2 + 24)(0)(1) = 0$ Right side: $a(2 - 8)(1) - b(2 - 8)(0) = a(-6)(1) - 0 = -6a$ So, $$0 = -6a \implies a = 0.$$ - Let $x=1$ (to eliminate $a$ term): At $x=1$: $$(1 - 1 + 6(1)^2)(1 - 2)(1 - 1) = a(1 - 1^3)(1 - 1) - b(1 - 1^3)(1 - 2)$$ Left side: $(1 - 1 + 6)(-1)(0) = 0$ Right side: $a(0)(0) - b(0)(-1) = 0$ This gives no information. 6. **Use another value, say $x=0$:** At $x=0$: $$(1 - 0 + 0)(0 - 2)(0 - 1) = a(0 - 0)(0 - 1) - b(0 - 0)(0 - 2)$$ Left side: $1 \times (-2) \times (-1) = 2$ Right side: $a(0)(-1) - b(0)(-2) = 0$ No information again. 7. **Rewrite the original equation with $a=0$ and solve for $b$: ** $$\frac{1 - x + 6x^2}{x - x^3} = - \frac{b}{x - 1}.$$ Multiply both sides by $(x - x^3)(x - 1)$: $$(1 - x + 6x^2)(x - 1) = -b(x - x^3).$$ Expand left side: $$(1 - x + 6x^2)(x - 1) = (1)(x - 1) - x(x - 1) + 6x^2(x - 1) = (x - 1) - (x^2 - x) + 6x^3 - 6x^2$$ Simplify: $$x - 1 - x^2 + x + 6x^3 - 6x^2 = 6x^3 - 7x^2 + 2x - 1.$$ Right side: $$-b(x - x^3) = -b x + b x^3 = b x^3 - b x.$$ 8. **Equate coefficients:** $$6x^3 - 7x^2 + 2x - 1 = b x^3 - b x.$$ Match powers of $x$: - For $x^3$: $6 = b$ - For $x^2$: $-7 = 0$ (contradiction) - For $x$: $2 = -b$ - For constant: $-1 = 0$ (contradiction) Contradiction means the original assumption $a=0$ is wrong or the form is not correct. 9. **Try partial fraction decomposition directly:** Rewrite denominator: $$x - x^3 = x(1 - x^2) = x(1 - x)(1 + x).$$ The right side has denominators $x-2$ and $x-1$, but the left denominator factors differently. 10. **Conclusion:** The given equation cannot hold for constants $a,b$ with denominators $x-2$ and $x-1$ because the denominators do not match the factorization of the left side. **Final answer:** $$a = 0, \quad b = 0$$ No such constants $a,b$ satisfy the equation as given.