Question

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

Answer:

$x^3 + 3x\²y + 3xy^2 + y$ $=283$

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

Step-by-step 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:

$x^3 + 3x\²y + 3xy^2 + y$ $=283$

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