Question

x cube + 3x square + 3x + 1 / 3 X + 1​

Answer 1

Step-by-step explanation:

I have tried to solve your problem thanku

Answer 2

Answer:

Step-by-step explanation:

Given :

To factorize :

$\frac{x^3+3x^2+3x+1}{3x+1}$

I thin it should be  x + 1 , not 3x + 1

Solution :

$\frac{x^3+3x^2+3x+1}{3x+1}$

⇒ $\frac{x^3+x^2  2x^2+2x + xx+1}{3x+1}$

⇒ $\frac{x^2(x+1)+2x(x+1) + 1(x+1)}{3x+1}$

⇒ $\frac{(x^2 + 2x + 1)(x+1)}{3x+1}$

⇒ $\frac{(x^2 + x + x + 1)(x+1)}{3x+1}$

⇒ $\frac{(x(x + 1) + 1(x + 1))(x+1)}{3x+1}$

⇒ $\frac{(x+1)(x+1)(x+1)}{3x+1}$

(or)

⇒ $\frac{(x+1)^3}{3x+1}$

If it was x + 1 , instead of 3x + 1,.

⇒ $\frac{(x+1)(x+1)(x+1)}{x+1} = (x+1)(x+1) = (x + 1)^2$