Question

factorise x cube + 3 X square + 3 x minus 7​

Answer 1

Answer:

Step-by-step explanation:

x^3 + 3x² +3x -7

= x^3 + 3x² + 3x +1 -8

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

Now, use identity

a^3 - b^3 = (a-b) ( a² + ab + b²)

We get, ( x+1 -2) { (x+1)² + 2(x+1) + 2² }

= (x-1) ( x² + 2x + 1 +2x + 2 + 4)

= (x-1) ( x² + 4x + 7)

Answer 2

The correct answer is $(x-1)(x^{2} +4x+7)$.

Given: The equation = $x^{3}+3x^{2} +3x-7$.

To Find: Factorize the equation.

Solution:

$x^{3}+3x^{2} +3x-7$

$x^{3}+3x^{2} +3x+1-8$

$(x^{3}+3x^{2} +3x+1)-8$

$(x+1)^{3} -2^{3}$  (equation)

Identity

$a^{3} -b^{3}=(a-b)(a^{2}+ab+b^{2} )$

By comparing both the equation and identity.

a = x+1

b = 2

$(x+1)^{3} -2^{3}=(x+1-2)((x+1)^{2}+2(x+1)+2^{2} )$

= $(x-1)(x^{2} +1+2x+2(x+1)+2^{2} )$

= $(x-1)(x^{2} +1+2x+2x+2+4 )$

= $(x-1)(x^{2} +4x+7)$

Hence, the factors of $x^{3}+3x^{2} +3x-7$ are $(x-1)(x^{2} +4x+7)$.

#SPJ2