Question

Find the value of 373 using the identity (x − y)3 = x3 − 3x2y + 3xy2 − y3. Show all work.

Hint: 373 = (40 − 3)3; therefore, x = 40 and y = 3.

Answer 1

`37^3 = 50653`

Explanation:

Given

`(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3`

Required

Find
`37^3`

Express 37 as 40 - 3

So, we have:

`37^3 = (40 - 3)^3`

Compare to
`(x -y)^3`

`x = 40\ and\ y = 3`

Substitute 40 for x and 3 for y in
`(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3`

`(40 - 3)^3 = 40^3 - 3*40^2*3 + 3*40*3^2 - 3^3`

Evaluate all exponents

`(40 - 3)^3 = 64000 - 3*1600*3 + 3*40*9 - 27`

Evaluate all products

`(40 - 3)^3 = 64000 - 14400 + 1080 - 27`

`(40 - 3)^3 = 50653`

Hence:

`37^3 = 50653`