Answer:
[tex]x^3 + 3x\²y + 3xy^2 + y[/tex] [tex]=283[/tex]
[tex]When\ x = 3\ and\ y = 4[/tex]
Step-by-step explanation:
Given
[tex]x^3 + 3x\²y + 3xy^2 + y[/tex]
Required
Solve when x = 3 and y = 4
To do this, we simply substitute 3 for x and 4 for y in [tex]x^3 + 3x\²y + 3xy^2 + y[/tex]
[tex]3^3 + 3 * 3^2 * 4 + 3 * 3 * 4^2 + 4[/tex]
[tex]27 + 3 * 9 * 4 + 3 * 3 * 16 + 4[/tex]
[tex]27 + 108 + 144+ 4[/tex]
[tex]283[/tex]
Hence:
[tex]x^3 + 3x\²y + 3xy^2 + y[/tex] [tex]=283[/tex]
[tex]When\ x = 3\ and\ y = 4[/tex]
