Question : If $a+b=1$, find the value of $a^{3}+b^{3} - ab-(a^{2}-b^{2})^{2}$.
Option 1: –1
Option 2: 1
Option 3: 0
Option 4: 2
Correct Answer: 0
Solution :
Given: $a+b=1$
We know that $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$
So, $a^{3}+b^{3}-ab-(a^{2}-b^{2})^{2}=(a+b)^{3}-3ab(a+b)-ab-(a+b)^{2}(a-b)^{2}$
Putting the value of $a+b=1$, we get,
$a^{3}+b^{3}-ab-(a^{2}–b^{2})^{2}$
$=(1)^{3}-3ab(1)-ab-(1)^{2}[(1)^{2}-4ab]$
$= 1-3ab-ab-1+4ab$
$=0$
Hence, the correct answer is 0.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
