How do you factor ${(b - c)}^{3} + {(c - a)}^{3} + {(a - b)}^{3}$ $(b−c)^3+(c−a)^3+(a−b)^3$ ?

Question

5983 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-01T19:00:59

$3 (b - c) (c - a) (a - b) .$ $3(b-c)(c-a)(a-b).$

If we use the following Result, we can immediately factorise the

given Exp. $= 3 (b - c) (c - a) (a - b) .$ $=3(b-c)(c-a)(a-b).$

Result : $x + y + z = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z .$ $x+y+z=0 rArr x^3+y^3+z^3=3xyz.$

Otherwise, consider the following :

Let, $u = b - c, v = c - a \Rightarrow u + v = b - a = - (a - b) .$ $u=b-c, v=c-a rArr u+v=b-a=-(a-b).$

Now, The Exp. $= u^{3} + v^{3} + {- (u + v)}^{3},$ $=u^3+v^3+{-(u+v)}^3,$

$= u^{3} + v^{3} - {(u + v)}^{3},$ $=u^3+v^3-(u+v)^3,$

$= u^{3} + v^{3} - {u^{3} + v^{3} + 3 u v (u + v)},$ $=u^3+v^3-{u^3+v^3+3uv(u+v)},$

$= - 3 u v (u + v) = 3 u v {- (u + v)},$ $=-3uv(u+v)=3uv{-(u+v)},$

$:." The Exp.="3(b-c)(c-a)(a-b).$

Enjoy Maths.!

How do you factor (b−c)^3+(c−a)^3+(a−b)^3(b−c)3+(c−a)3+(a−b)3(b−c)^3+(c−a)^3+(a−b)^3?