Question

88. Rút gọn biểu thức:

\(A=\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

Answer 1

Tử \(a^3+b^3+c^3-3abc\)

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

\(=(a+b)^3-3ab(a+b)+c^3-3abc\)

\(=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)\)

\(=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)\)

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

Khi đó \(A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}=a+b+c\)

Answer 2

Ta có :

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

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\) Thay vào biểu thức ta được:

\(A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)