Question

Pls help me.proove it ^​

Answer 1

Answer:

We verified that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved

Step-by-step explanation:

Given equation is $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

We have to prove that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

That is to prove that LHS=RHS

Now taking RHS

$\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

$=\frac{a+b+c}{2}[a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2]$  (using $(a-b)^2=a^2-2ab+b^2$)

$=\frac{a+b+c}{2}[2a^2-2ab+2b^2-2bc+2c^2-2ac]$  (adding the like terms)

$=\frac{a+b+c}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ac]$

$=\frac{a+b+c}{2}\times 2[a^2+b^2+c^2-ab-bc-ac]$

$=a+b+c[a^2+b^2+c^2-ab-bc-ac]$

Now multiply the each term to another each term in the factor

$=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba62+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2]$

$=a^3+b^3+c^3-3abc$ (adding the like terms and other terms getting cancelled)

$=a^3+b^3+c^3-3abc$ =LHS

Therefore LHS=RHS

Therefore $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved.