(7 + 1/2)³ × (7 - 1/2)⁴ / (2 + 1/7)³ × (2 - 1/7)⁴ equals to

(7 + 1/2)³ = (15/2)³ = 3375/8

(7 - 1/2)⁴ = (13/2)⁴ = 28561/16

(2 + 1/7)³ = (15/7)³ = 3375/343

(2 - 1/7)⁴ = (13/7)⁴ = 28561/2401

Now, we can substitute these values back into the expression:

(3375/8) * (28561/16) / (3375/343) * (28561/2401)

= (3375 * 28561) / (8 * 16) * (343 * 2401)

= 96309375 / 2744

= 35125

Therefore, (7 + 1/2)³ * (7 - 1/2)⁴ / (2 + 1/7)³ * (2 - 1/7)⁴ equals to 35125.

(7 1/2)³ × (7 - 1/2)⁴ / (2 1/7)³ × (2 - 1/7)⁴

$$\text{Let }x=7\frac{1}{2}\text{ and }y=2\frac{1}{7}$$

$$=(x)³×(x-1)⁴/(y)³×(y-1)⁴$$

$$=x³(x-1)⁴/y³(y-1)⁴$$

$$=x³×\frac{(x-1)}{(y-1)}×\frac{(x-1)³}{(y-1)³}$$

$$=x³×\frac{(x-1)}{(y-1)}×\frac{(x³-3x²+3x-1)}{(y³-3y²+3y-1)}$$

$$=\frac{x⁴-3x³+3x²-x}{y³-3y²+3y-1}$$

$$=\frac{x(x³-3x²+3x-1)}{y³-3y²+3y-1}$$

Putting values of x and y

$$=(7\frac{1}{2})×\frac{(7\frac{1}{2})³-3(7\frac{1}{2})²+3(7\frac{1}{2})-1}{(2\frac{1}{7})³-3(2\frac{1}{7})²+3(2\frac{1}{7})-1}$$

$$=(7\frac{1}{2})×\frac{(35³/8-3×315/8+3×77/8-1/8)}{(343/343-3×81/343+3×9/343-1/343)}$$

$$=(7\frac{1}{2})×\frac{(105-945+231-1)/8}{(343-243+27-1)/343}$$

$$=(7\frac{1}{2})×\frac{85/8}{122/343}$$

$$=(7\frac{1}{2})×\frac{343×85}{8×122}$$

$$=(7\frac{1}{2})×\frac{29105}{976}$$

$$=(7\frac{1}{2})×29.828$$

$$=213.7$$