The value of $$\frac{(2^3 - 1)(3^3 - 1).......(100^3 - 1)}{(2^3 + 1)(3^3 + 1).......(100^3 + 1)}$$ is closest to
Solution
$$a^3-1$$ could be written as (a-1)($$a^2+ab+b^2$$) and $$a^3+1$$ could be written as (a+1)($$a^2-ab+b^2$$).
So, the expression could be written as: $$\frac{\left(2-1\right)\left(3-1\right)\left(4-1\right)...}{\left(2+1\right)\left(3+1\right)\left(4+1\right)...}\times\ \frac{\left(2^2+2+1^2\right)\left(3^2+3+1^2\right)...}{\left(2^2-2+1^2\right)\left(3^2-3+1^2\right)...}$$.
The first fraction will be: $$\frac{1\cdot2\cdot3...99}{3\cdot4\cdot5...101}=\frac{1\cdot2}{100\cdot101}=\frac{1}{50\cdot101}$$.
For the 2nd part, $$a^2+a+1^2=\left(a+1\right)^2-\left(a+1\right)+1^2$$. This means that $$2^2+2+1^2=\left(3\right)^2-\left(3\right)+1^2$$.
So, the 2nd fraction could be written as: $$\frac{7\cdot13\cdot21...}{3\cdot7\cdot13...}=\frac{100^2+100+1^2}{3}=\frac{10101}{3}$$.
The product of the 2 fractions will be : $$\frac{1}{50\cdot101}\times\ \frac{10101}{3}=0.667\ \approx\ \frac{2}{3}$$
