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Sagot :
To find the product of the polynomials \((a + b + c)(a² + b² + c² + ab - bc - ca)\), we can use the distributive property (also known as the FOIL method for binomials). Here's the step-by-step process:
1. Distribute each term in the first polynomial \((a + b + c)\) to each term in the second polynomial \((a² + b² + c² + ab - bc - ca)\).
Let's break it down:
\[
(a + b + c)(a² + b² + c² + ab - bc - ca)
\]
First, distribute \(a\):
\[
a(a² + b² + c² + ab - bc - ca) = a³ + ab² + ac² + a²b - abc - a²c
\]
Next, distribute \(b\):
\[
b(a² + b² + c² + ab - bc - ca) = ba² + b³ + bc² + ab² - b²c - bac
\]
Finally, distribute \(c\):
\[
c(a² + b² + c² + ab - bc - ca) = ca² + cb² + c³ + cab - cbc - c²a
\]
Now, combine all the terms:
\[
a³ + ab² + ac² + a²b - abc - a²c + ba² + b³ + bc² + ab² - b²c - bac + ca² + cb² + c³ + cab - cbc - c²a
\]
Combine like terms:
\[
a³ + b³ + c³ + 2ab² + 2ac² + 2a²b + 2bc² + 2cab - abc - b²c - cbc - a²c - bac - c²a
\]
This is the expanded form of the product of the given polynomials.
1. Distribute each term in the first polynomial \((a + b + c)\) to each term in the second polynomial \((a² + b² + c² + ab - bc - ca)\).
Let's break it down:
\[
(a + b + c)(a² + b² + c² + ab - bc - ca)
\]
First, distribute \(a\):
\[
a(a² + b² + c² + ab - bc - ca) = a³ + ab² + ac² + a²b - abc - a²c
\]
Next, distribute \(b\):
\[
b(a² + b² + c² + ab - bc - ca) = ba² + b³ + bc² + ab² - b²c - bac
\]
Finally, distribute \(c\):
\[
c(a² + b² + c² + ab - bc - ca) = ca² + cb² + c³ + cab - cbc - c²a
\]
Now, combine all the terms:
\[
a³ + ab² + ac² + a²b - abc - a²c + ba² + b³ + bc² + ab² - b²c - bac + ca² + cb² + c³ + cab - cbc - c²a
\]
Combine like terms:
\[
a³ + b³ + c³ + 2ab² + 2ac² + 2a²b + 2bc² + 2cab - abc - b²c - cbc - a²c - bac - c²a
\]
This is the expanded form of the product of the given polynomials.
Answer:
a³ + b³ + c³ - abc + 2ab(a + b)
Step-by-step explanation:
(a + b + c)(a² + b² + c² + ab - bc - ca)
= a³ + ab² + ac² + a²b - abc - a²c +
a²b + b³ + bc² + ab² - b²c - abc +
a²c + b²c + c³ + abc - bc² - ac²
= a³ + b³ + c³ + 2ab² + 2a²b - abc
= a³ + b³ + c³ - abc + 2ab(a + b)
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