Consider the equation (1+a+b)^2=3(1+a^2+b^2) | vedclass

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(D) Given equation: $(1+a+b)^2=3(1+a^2+b^2)$Expanding the left side: $1+a^2+b^2+2a+2b+2ab = 3+3a^2+3b^2$Rearranging the terms: $2a^2+2b^2-2a-2b-2ab+2=

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there is exactly one solution pair $(a, b)$

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(D) Given equation: $(1+a+b)^2=3(1+a^2+b^2)$Expanding the left side: $1+a^2+b^2+2a+2b+2ab = 3+3a^2+3b^2$Rearranging the terms: $2a^2+2b^2-2a-2b-2ab+2=0$Dividing by $2$: $a^2+b^2-a-b-ab+1=0$Multiplying by $2$ again to complete squares: $2a^2+2b^2-2a-2b-2ab+2=0$This can be rewritten as: $(a^2-2a+1)+(b^2-2b+1)+(a^2+b^2-2ab)=0$$(a-1)^2+(b-1)^2+(a-b)^2=0$Since the sum of squares of real numbers is zero if and only if each term is zero:$a-1=0, b-1=0, a-b=0$This implies $a=1$ and $b=1$.Thus,there is exactly one solution pair $(a, b) = (1, 1)$.

Consider the equation $(1+a+b)^2=3(1+a^2+b^2)$ where $a, b$ are real numbers. Then,

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