If \(a+b=1\), then find the value of \(a^4+b^4-a^3-b^3-2a^2b^2+ab\).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
The terms group into \(a^4+b^4-2a^2b^2=(a^2-b^2)^2\) and \(a^3+b^3\). Both can be written with \(a+b\) and \(ab\) only, and then \(a+b=1\) is put in.
\(a^4+b^4-2a^2b^2=(a^2-b^2)^2=(a+b)^2(a-b)^2=(a-b)^2\)
\((a-b)^2=(a+b)^2-4ab=1-4ab\)
\(a^3+b^3=(a+b)^3-3ab(a+b)=1-3ab\)
Expression \(=(1-4ab)-(1-3ab)+ab=1-4ab-1+3ab+ab=0\)