Question

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

Answer 1

`a = 1, b = 1`

Explanation:

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

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

Explanation:

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)`