Question

Prove: a^3+b^3+c^3-3abc=1/2(a+b+c)

Answer 1

Solution:

For verifying the given, we need to prove that the left hand side (L.H.S) of the equation is equal to right hand side (R.H.S).

Therefore R.H.S,

$\small \bold{= \frac{1}{2}\left((a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)\right) \ldots \ldots \ldots}$

By using the formula

$(a-b)^{2}=a^{2}+b^{2}+2 a b$

Then, as per the above formula the equation 1 becomes,

$\small \bold{=\frac{1}{2}(a+b+c)\left(a^{2}-2 a b+b^{2}\right)+\left(b^{2}-2 b c+c^{2}\right)+\left(c^{2}-2 c a+a^{2}\right)}$

$\small \bold{=\frac{1}{2}(a+b+c) 2\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)}$

$\small\bold{=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)}$

$\begin{gathered}\begin{aligned}=& a^{3}+a b^{2}+a c^{2}-b a^{2}-a b c-c a^{2}+b a^{2}+b^{3}+b c^{2}-a b^{2}-b^{2} c-c a b+c a^{2} \\ &+c b^{2}+c^{3}-a b c-b c^{2}-c^{2} a \end{aligned}\end{gathered}$

After simplification,

$=a^{3}+b^{3}+c^{3}-3 a b c$

L.H.S=R.H.S

Hence, the given equation is proved, that the “left hand side” of the equation is equal to the “right hand side”.