Question

C.
Find the value of x3 + 3x²y + 3xy2 + y when x = 3 and y = 4.
涨涨涨涨
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Answer 1

`x^3 + 3x\²y + 3xy^2 + y`
`=283`

`When\ x = 3\ and\ y = 4`

Explanation:

Given

`x^3 + 3x\²y + 3xy^2 + y`

Required

Solve when x = 3 and y = 4

To do this, we simply substitute 3 for x and 4 for y in
`x^3 + 3x\²y + 3xy^2 + y`

`3^3 + 3 * 3^2 * 4 + 3 * 3 * 4^2 + 4`

`27 + 3 * 9 * 4 + 3 * 3 * 16 + 4`

`27 + 108 + 144+ 4`

`283`

Hence:

`x^3 + 3x\²y + 3xy^2 + y`
`=283`

`When\ x = 3\ and\ y = 4`