Question

a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca), if a+b+c=0 then prove that (1). a^3+b^3+c^3= 3abc (2). a^2/bc+b^2/ca+c^2/ab=3

Answer 1

Answer:

We know that any thing multiplied by zero becomes a zero

So if (a+b+c) = 0 the RHS becomes zero

So if we shift 3abc on the other side of becomes

a(cube)+b(cube)+(c)cube = 3abc.......(1)

Now for the second part just divide this equation that is equation 1 with abc you will get the answer