If `x=3\ a n d\ y=-1,` find the values of each of the using identity: `(x/4-y/3)\ ((x^2)/(16)+(x y)/(12)+(y^2)/9)`

Explanation

we know that,
`a^3-b^3=(a-b)(a^2+ab+b^2)`
using this formula,
we can write it as,
`(x/4-y/3)\ ((x^2)/(16)+(x y)/(12)+(y^2)/9)`
`=(x/4)^3-(y/3)^3`
now putting the values of xand y from the question,
we get,
`=(3/4)^3-(-1/3)^3`
`=27/64+1/27`
`=729/1728`