The equation $$(x+1)^3-(x-1)^3=0$$ is :

Correct option is C

Given:

​$$(x+1)^3 - (x-1)^3 = 0$$  

Formula used:

​$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ 

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

Solution: 

​ $$(x+1)^3 - (x-1)^3 = 0$$​​

​$$(x+1)^3 - (x-1)^3 =0 \\ \ \\(x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) = 0$$​ 

​$$x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 = 0$$ 

$$6x^2 + 2 = 0$$ 

$$6x^2 = -2$$ 

​$$x^2 = -\frac{1}{3}$$ 

$$x = \pm \sqrt{-\frac{1}{3}}$$ 

Since the solutions involve the imaginary unit , the roots are non-real.

The correct answer is Option C :  quadratic equation with non-real roots.

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