a3 +b3 +c3 -3abc factorise
Answers
Answer:
) Let f = a³(b - c) + b³(c - a) + c³(a - b)
ii) Put a = b, in the above; then f = b³(b - c) + b³(c - b) + c³(b - b) = b³(b - c) - b³(b - c) + 0 = 0
So by factor theorem (a - b) is one factor of f.
Similarly it can be shown that, both (b - c) and (c - a) are the other two factors.
Hence, f = (a - b)(b - c)(c - a)*{k(a + b + c)}, where k is a constant.
[Why an additional factor of k(a + b + c)?
The given is a homogeneous polynomial of 4th degree. (On expanding sum of exponents of each term is 4, so it is a homogeneous one in 4th degree). But (a - b)(b - c)(c - a) results only in homogenous 3rd degree; so as to make 4th degree, it is multiplies by k(a + b + c)]
iii) In the above we need to evaluate the value of k.
For which put a = 1, b = 2 and c = 3 and evaluate;
==> (1 - 2)(2 - 3)(3 - 1)*k(1 + 2 + 3) = 1³(2 - 3) + 2³(3 - 1) + 3³(1 - 2)
==> 12k = -12; solving k = -1
Thus f = -(a - b)(b - c)(c - a)(a + b + c)