If x is a rational number and frac{(x+1)^{3}- | vedclass

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(B) Given the equation: $\frac{(x+1)^{3}-(x-1)^{3}}{(x+1)^{2}-(x-1)^{2}}=2$Using algebraic identities $(a+b)^3 - (a-b)^3 = 2b^3 + 6a^2b$ and $(a+b)^2

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$4$

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(B) Given the equation: $\frac{(x+1)^{3}-(x-1)^{3}}{(x+1)^{2}-(x-1)^{2}}=2$Using algebraic identities $(a+b)^3 - (a-b)^3 = 2b^3 + 6a^2b$ and $(a+b)^2 - (a-b)^2 = 4ab$:Numerator: $(x+1)^3 - (x-1)^3 = 2(1)^3 + 6(x)^2(1) = 2 + 6x^2$Denominator: $(x+1)^2 - (x-1)^2 = 4(x)(1) = 4x$Substituting these into the equation:$\frac{6x^2 + 2}{4x} = 2$Simplify the fraction by dividing by $2$:$\frac{3x^2 + 1}{2x} = 2$Multiply both sides by $2x$:$3x^2 + 1 = 4x$Rearrange into a standard quadratic equation:$3x^2 - 4x + 1 = 0$Factor the quadratic equation:$(3x - 1)(x - 1) = 0$This gives two possible values for $x$: $x = 1/3$ or $x = 1$.If $x = 1/3$,the sum of the numerator and denominator is $1 + 3 = 4$.If $x = 1$,the sum of the numerator and denominator is $1 + 1 = 2$ (not in options).Thus,the correct value is $4$.

If $x$ is a rational number and $\frac{(x+1)^{3}-(x-1)^{3}}{(x+1)^{2}-(x-1)^{2}}=2$,then the sum of the numerator and denominator of $x$ is

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