#(1+a+b)^2 =3(1+a^2+b^2)# Let's do it???

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Answer 1 · 2017-08-11T12:18:33

#a = 1, b = 1#

Solving the traditional way

#(1 + a + b)^2 - 3 (1 + a^2 + b^2) = 0 rArr 1 - a + a^2 - b - a b + b^2=0#

Now solving for #a#

#a = 1/2 (1 + b pm sqrt[3] sqrt[2 b - b^2-1])# but #a# must be real so the condition is

#2 b - b^2-1 ge 0# or #b^2-2b+1 le 0 rArr b = 1#

now substituting and solving for #a#

#1 - 2 a + a^2 = 0 rArr a = 1# and the solution is

#a = 1, b = 1#

Another way to do the same

#(1 + a + b)^2 - 3 (1 + a^2 + b^2) = 0 rArr 1 - a + a^2 - b - a b + b^2=0#

but

# 1 - a + a^2 - b - a b + b^2 = (a-1)^2+(b-1)^2-(a-1)(b-1)#

and concluding

#(a-1)^2+(b-1)^2-(a-1)(b-1) = 0 rArr a = 1, b= 1#

Answer 2 · 2017-08-11T20:24:22

D. There is exactly one solution pair #(a, b) = (1, 1)#

Given:

#(1+a+b)^2 = 3(1+a^2+b^2)#

Note that we can make this into a nice symmetric homogeneous problem by generalising to:

#(a+b+c)^2 = 3(a^2+b^2+c^2)#

then set #c=1# at the end.

Expanding both sides of this generalised problem, we have:

#a^2+b^2+c^2+2ab+2bc+2ca = 3a^2+3b^2+3c^2#

Subtracting the left hand side from both sides, we get:

#0 = 2a^2+2b^2+2c^2-2ab-2bc-2ca#

#color(white)(0) = a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2#

#color(white)(0) = (a-b)^2+(b-c)^2+(c-a)^2#

For real values of #a#, #b# and #c#, this can only hold if all of #(a-b)#, #(b-c)# and #(c-a)# are zero and hence:

#a = b = c#

Then putting #c=1# we find the only solution to the original problem, namely #(a, b) = (1, 1)#