Question

5. Factorize
(c) x^3-1+3/x-1/x^3​

Answer 1

To factorize x^3 + 1/x^3 - 2, take LCM of the expression.

Expression becomes, (x^6 - 2 x^3 + 1) / x^3. Now numerator is easily factorisable as (x^3 - 1)^2

So final expression is (x^3 - 1)^2 / x^3.

Edit-1: As requested by Rishi Raj, for further factorization with real coefficients, we can use

x^3 - 1 = (x - 1)*(x^2 + x + 1)

So (x^3 - 1)^2 / x^3 = [(x - 1)^2 * (x^2 + x + 1)^2] / x^3

This cannot be factorized further with real coefficients. However if complex coefficients are allowed, we include imaginary cube roots of unity.

i.e.,

x^2 + x + 1 = (x + w)(x + w^2), where

w = [1 + i√3]/2

Hope it helped !..

Answer 2

Answer:

As we see that the x has a power of 3 so we begin by cubbing the whole equation.

So we start solving like:

[math]x + \dfrac{1}{x} = 3[/math]

[math](x + \dfrac{1}{x})^3 = 3^3[/math]

[math]x^3 + 3.x^2.\dfrac{1}{x} + 3.x.\dfrac{1}{x^2} + \dfrac{1}{x^3} = 27[/math]

[math](x^3 + \dfrac{1}{x^3}) + 3.x^2.\dfrac{1}{x} + 3.x.\dfrac{1}{x^2} = 27[/math]

[math]x^3 + \dfrac{1}{x^3} + (3x + \dfrac{3}{x}) = 27[/math]

[math]x^3 + \dfrac{1}{x^3} + 3(x + \dfrac{1}{x}) = 27[/math]

[math]x^3 + \dfrac{1}{x^3} + 3 × 3 = 27[/math]

[math]x^3 + \dfrac{1}{x^3} + 9 = 27[/math]

[math]x^3 + \dfrac{1}{x^3} = 18[/math]

Therefore the answer is 18