Answer:
We verified that [tex]a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2][/tex]
Hence proved
Step-by-step explanation:
Given equation is [tex]a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2][/tex]
We have to prove that [tex]a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2][/tex]
That is to prove that LHS=RHS
Now taking RHS
[tex]\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2][/tex]
[tex]=\frac{a+b+c}{2}[a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2][/tex] (using [tex](a-b)^2=a^2-2ab+b^2[/tex])
[tex]=\frac{a+b+c}{2}[2a^2-2ab+2b^2-2bc+2c^2-2ac][/tex] (adding the like terms)
[tex]=\frac{a+b+c}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ac][/tex]
[tex]=\frac{a+b+c}{2}\times 2[a^2+b^2+c^2-ab-bc-ac][/tex]
[tex]=a+b+c[a^2+b^2+c^2-ab-bc-ac][/tex]
Now multiply the each term to another each term in the factor
[tex]=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba62+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2][/tex]
[tex]=a^3+b^3+c^3-3abc[/tex] (adding the like terms and other terms getting cancelled)
[tex]=a^3+b^3+c^3-3abc[/tex] =LHS
Therefore LHS=RHS
Therefore [tex]a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2][/tex]
Hence proved.
