Question

cho a+b+c=0;chứng minh rằng a³+a²c-abc+b²c+b³=0

Answer 1

Solution:

\(a^3+a^2c-abc+b^2c+b^3\)

\(=a^2\left(a+c\right)+b^2\left(b+c\right)-abc\)

\(=a^2\cdot\left(-b\right)+b^2\cdot\left(-a\right)-abc\)

\(=-ab\left(a+b+c\right)\)

\(=0\)

Answer 2

Ta có:

\(a^3+a^2c-abc+b^2c+b^3=\left(a^3+b^3\right)+\left(a^2c-abc+b^2c\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)

\(=\left(a+b+c\right)\left(a^2-ab+b^2\right)=0\)

Vì \(a+b+c=0\) nên \(a^3+a^2c-abc+b^2c+b^3=0\)

Answer 3

\(a^3+a^2c-abc+b^2c+b^3=0\\ \Leftrightarrow\left(a^3+b^3\right)+\left(a^2c-abc+b^3c\right)=0\\ \Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)=0\\ \Leftrightarrow\left(a^2-ab+b^2\right)\left(a+b+c\right)=0\left(luônđúngvìa+b+c=0\right)\)

Study well!!