Context: If a³+b³+c³–3abc=0, then find the value of (a²/bc+b²/ac–3).(a) –c²/ab(b) –c²/bc(c) –c³/ba(d) –c/a Answer question

The given equation is a³+b³+c³–3abc=0. This is a well-known identity which states that if a + b + c = 0, then a³+b³+c³–3abc=0.

So, from the given equation, we can infer that a + b + c = 0.

Now, let's find the value of (a²/bc+b²/ac–3).

Substitute a = -(b + c) into the equation:

= [(-b-c)²/bc + b²/(-b-c)c - 3] = [b²/bc + 2bc/bc + c²/bc + b²/ac - 3] = [1 + 2 - c/b - 3] = 0

So, the value of (a²/bc+b²/ac–3) is 0.

Now, let's find the value of the options:

(a) –c²/ab = -c²/-(b+c) = c²/(b+c) = c²/c = c (since b + c = c) (b) –c²/bc = -c (since b + c = c) (c) –c³/ba = -c³/-c = c² (since a = -c) (d) –c/a = -c/-c = 1 (since a = -c)

So, none of the options match the value of (a²/bc+b²/ac–3).