How do you rationalize the denominator and simplify 1/(sqrta+sqrtb+sqrtc)1√a+√b+√c?
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))(−√a+√b+√c)(√a−√b+√c)(√a+√b−√c)
Explanation:
For brevity, write:
alpha = sqrt(a)α=√a
beta = sqrt(b)β=√b
gamma = sqrt(c)γ=√c
Multiply numerator and denominator by:
(-alpha+beta+gamma)(alpha-beta+gamma)(alpha+beta-gamma)(−α+β+γ)(α−β+γ)(α+β−γ)
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−α3−β3−γ3+α2β+β2γ+αγ2+αβ2+βγ2+α2γ−2αβγ
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)(α+β+γ)(−α3−β3−γ3+α2β+β2γ+αγ2+αβ2+βγ2+α2γ−2αβγ)
=-alpha^4-beta^4-gamma^4+2alpha^2beta^2+2beta^2gamma^2+2alpha^2gamma^2=−α4−β4−γ4+2α2β2+2β2γ2+2α2γ2
=2ab+2bc+2ac-a^2-b^2-c^2=2ab+2bc+2ac−a2−b2−c2
So we can write:
1/(sqrt(a)+sqrt(b)+sqrt(c))1√a+√b+√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)=(−√a+√b+√c)(√a−√b+√c)(√a+√b−√c)2ab+2bc+2ac−a2−b2−c2
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)=(b+c−a)√a+(a+c−b)√b+(a+b−c)√c−2√a√b√c2ab+2bc+2ac−a2−b2−c2
