find the value of x³+3x²y+y³if x=2and y=3
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Answer:
71
Step-by-step explanation:
x^3+3x^2y+y^3
2^3+3×2^2×3+3^3
8+36+27
71
Answer:
x³+3x²y+y³
= by putting values x=2 and y=3
= 2³+3*(2²*3)+3³
=8+36+27
= 71
Step-by-step explanation:
hope it will help you