Question

verify a3+b3+c3-3abc=1/2(a+b+c) {(a-b)2+(b-c)2+(c-a)2

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,  

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

By using the formula $(a-b)^{2}=a^{2}+b^{2}+2 a b$

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

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

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

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

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

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”.

Answer 2

Solution:

$Given\\a^{3}+b^{3}+c^{3}-3abc=\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$

$R.H.S=\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$

$=\frac{1}{2}(a+b+c)[a^{2}+b^{2}-2ab+b^{2}+c^{2}-2bc+c^{2}+a^{2}-2ca]$

$=\frac{1}{2}(a+b+c)[2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca]$

$=\frac{1}{2}(a+b+c)[2(a^{2}+b^{2}+c^{2}-ab-bc-ca)]$

$=\frac{1}{2}(a+b+c)(2)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

$=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

$=a^{3}+b^{3}+c^{3}-3abc\\=L.H.S$

Hence , Proved.

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