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
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
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.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : Directions: If 1 * 2 = 1, 2 * 3 = –1 and 3 * 4 = –5, then find the value of 7 * 9 = ?
Question : If $a^{2}+1=a$, the value of $a^{3}$ is:
Question : If $(a+\frac{1}{a})=–2$, then the value of $a^{1000}+a^{–1000}$ is:
Question : If $a+\frac{1}{a-2}=4$, then the value of $(a-2)^{2}+(\frac{1}{a-2})^{2}$ is:
Question : If the sum of the roots of a quadratic equation is 1 and the product of the roots is –20, find the quadratic equation.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile