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