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(B) Given equation: $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$Taking the common denominator on the left side:$\frac{(x+5)-(x-3)}{(x-3)(x+5)}=\frac{1}{6

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$-9$ and $7$

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(B) Given equation: $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$Taking the common denominator on the left side:$\frac{(x+5)-(x-3)}{(x-3)(x+5)}=\frac{1}{6}$Simplifying the numerator and denominator:$\frac{x+5-x+3}{x^{2}+5x-3x-15}=\frac{1}{6}$$\frac{8}{x^{2}+2x-15}=\frac{1}{6}$Cross-multiplying:$x^{2}+2x-15 = 48$$x^{2}+2x-15-48 = 0$$x^{2}+2x-63 = 0$Factoring the quadratic equation:$x^{2}+9x-7x-63 = 0$$x(x+9)-7(x+9) = 0$$(x+9)(x-7) = 0$Setting each factor to zero:$x+9 = 0 \implies x = -9$$x-7 = 0 \implies x = 7$Thus,the roots of the equation are $-9$ and $7$.

Solve the following equation using the method of factorization: $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6} \,(x \neq 3, -5)$

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