Let a, b, c be positive real numbers such that logab+logbc+logca=0.Find (logab)3+(logbc)3+(logca)3.
+130466 Using the change of base rule
Let x = log b/ log a y = log c / log b z = log a / log c
x + y + z = 0
x + y = -z square both sides
x^2 + y^2 + 2xy = z^2
x^2 + y^2 = z^2 - 2xy
x^3 + y^3 + z^3 =
(x + y) ( x^2 - xy + y^2) + z^3 =
(-z) ( z^2 - 2xy - xy) + z^3 =
(-z) ( z^2 - 3xy) + z^3 =
-z^3 + 3xyz + z^3 =
3xyz =
3 [ log b / log a ] [ log c / log b ] [ log a / log c] =
3 [ log a/ loga ] [ log b /log b ] [ log c / log c ] =
3 [ 1] [ 1 ] [ 1 ] =
3
