How do you rationalize the denominator and simplify #1/(sqrta+sqrtb+sqrtc)#?

1 Answer
Sep 27, 2015

Answer:

Multiply numerator and denominator by:

#(-sqrt(a)+sqrt(b)+sqrt(c))(sqrt(a)-sqrt(b)+sqrt(c))(sqrt(a)+sqrt(b)-sqrt(c))#

Explanation:

For brevity, write:

#alpha = sqrt(a)#
#beta = sqrt(b)#
#gamma = sqrt(c)#

Multiply numerator and denominator by:

#(-alpha+beta+gamma)(alpha-beta+gamma)(alpha+beta-gamma)#

Multiplied out in full, that is:

#-alpha^3-beta^3-gamma^3+alpha^2beta+beta^2gamma+alphagamma^2+alphabeta^2+betagamma^2+alpha^2gamma-2alphabetagamma#

The denominator is:

#(alpha+beta+gamma)(-alpha^3-beta^3-gamma^3+alpha^2beta+beta^2gamma+alphagamma^2+alphabeta^2+betagamma^2+alpha^2gamma-2alphabetagamma)#

#=-alpha^4-beta^4-gamma^4+2alpha^2beta^2+2beta^2gamma^2+2alpha^2gamma^2#

#=2ab+2bc+2ac-a^2-b^2-c^2#

So we can write:

#1/(sqrt(a)+sqrt(b)+sqrt(c))#

#=((-sqrt(a)+sqrt(b)+sqrt(c))(sqrt(a)-sqrt(b)+sqrt(c))(sqrt(a)+sqrt(b)-sqrt(c)))/(2ab+2bc+2ac-a^2-b^2-c^2)#

or if you prefer:

#=((b+c-a)sqrt(a)+(a+c-b)sqrt(b)+(a+b-c)sqrt(c)-2sqrt(a)sqrt(b)sqrt(c))/(2ab+2bc+2ac-a^2-b^2-c^2)#