Question : If $a+b=1$, then $a^4+b^4-a^3-b^3-2a^2b^2+ab$ is equal to:
Option 1: 1
Option 2: 2
Option 3: 4
Option 4: 0
Correct Answer: 0
Solution :
Given: $a+b=1$ (equation 1)
We know that the algebraic identities are $(a-b)^2=a^2+b^2-2ab$, $(a^3+b^3)=(a+b)(a^2-ab+b^2)$ and $(a^2-b^2)=(a+b)(a-b)$.
So, $a^4+b^4-2a^2b^2-a^3-b^3+ab$
$=(a^2-b^2)^2-(a^3+b^3)+ab$
$=(a+b)^2(a-b)^2-(a+b)(a^2-ab+b^2)+ab$
Substituting the value from equation 1 in the above expression we get,
$a^4+b^4-2a^2b^2-a^3-b^3+ab=(a-b)^2-a^2+ab-b^2+ab$
$=(a-b)^2-(a^2-2ab+b^2)$
$=(a-b)^2-(a-b)^2=0$
Hence, the correct answer is 0.
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