Question

no spamming please prove it​

Answer 1

Answer:

Step-by-step explanation:

Mark it as the brainiest one please

Answer 2

Required to prove :-

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

We start from the L.H.S and we will try to prove that L.H.S = R.H.S .

$\mathsf{a^3+b^3+c^3-3abc}\\\\\textbf{We know that by expansion a^3+b^3+c^3-3abc can be written as :}\\\\\implies \mathsf{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}\\\\\implies \mathsf{\dfrac{1}{2}(a+b+c)2(a^2+b^2+c^2-ab-bc-ac)}\\\\\implies \mathsf{\dfrac{1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ac)}\\\\\implies \mathsf{\dfrac{1}{2}(a+b+c)(a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ac)}\\\\\implies \mathsf{\dfrac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]}$

Hence Proved .

NOTE :

The formulas that were used in the above proof are :-

$\bigstar \mathsf{(x-y)^2=x^2+y^2-2xy}\\\\\bigstar \mathsf{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}$