Question

 cmr

c) (a+b+c)³ -a ³ -b ³ -c ³=3(a+b)(b+c)(c+a)

d) a³+b³+c³ -3abc=(a+b+c)(a²+b² +c² -ab-bc-ca)

e) (a+b+c)³ -(b+c-a)³ -(a+c-b) ³ -(a+b-c)³=24abc

 

Answer 1

d) Ta có: \(a^3+b^3+c^3-3abc\)

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

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

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

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