Question

Chứng minh nếu a+b+c=5 thì \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}=5\)

Answer 1

\(\dfrac{a^3+b^3+c^3-3bac}{a^2+b^2+c^2-ab-ac-bc}\)

\(=\dfrac{\left(a+b\right)^3+c^3-3ba\left(a+b\right)-3bac}{a^2+b^2+c^2-ab-ac-bc}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-ac-bc}\)

=a+b+c

=5