The value of $$\frac{1-x^4}{1+x}÷\frac{1+x^2}{x}×\frac{1}{x(x-1)}$$​ is:

(if x is not equal to 0, –1, 1)

Correct option is D

Given:

​$$\frac{1 - x^4}{1 + x} \div \frac{1 + x^2}{x} \times \frac{1}{x(x - 1)}$$​

Concept Used:

a²- b²=(a - b) (a + b)

​Solution:

$$\frac{1 - x^4}{1 + x} \div \frac{1 + x^2}{x} \times \frac{1}{x(x - 1)}$$​​

$$=\frac{(1+x)(1 - x)(1 + x^2)}{(1+x)} \div \frac{1 + x^2}{x} \times \frac{1}{x(x - 1)}$$​

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

$$=(1 - x) \times x \times \frac{1}{x(x - 1)}$$​

$$=(- (x - 1)) \times x \times \frac{1}{x(x - 1)}$$​

= −1