Question

factorize (x+1)^3-(x-1)^3 ​

Answer 1

$(x + 1) {}^{3} - (x - 1) {}^{3} = 0\\ (x {}^{3} + 1 + 3x {}^{2} + 3x) - (x {}^{3} - 1 - 3x {}^{2} + 3x) = 0\\ x {}^{3} + 1 + 3x {}^{2} + 3x - x {}^{3} + 1 + 3x {}^{2} - 3x = 0\\ 6x {}^{2} + 2 = 0 \\ 6x {}^{2} = - 2 \\ x {}^{2} = - \frac{2}{6} \\ x = - \sqrt{ \frac{2}{6} } \\ x = - \sqrt{ \frac{1}{3} }$

Answer 2

let X+1=a

and x-1=b

(a-b)^3=a^3 - b^3 +3b^2a - 3a^2b

a^3 - b^3=(a - b)^3+3ab(a - b)

a^3 - b^3=(a - b)^2(a^2+ab+b^2)

now put the value of a and b

(x+1)^3 - b^3(let z)=8((x+1)^2+(x-1)(x+1)+(x-1)^2

z=8(3x^2+1)

$z = 8(( { \sqrt{3} x})^{2} - {i}^{2} )$

$z = 8(( \sqrt{3} - i)( \sqrt{3} +i)$