Question : If $(x–4)(x^2+4x+16)=x^3–p$, then $p$ is equal to:
Option 1: 27
Option 2: 8
Option 3: 64
Option 4: 0
Latest: SSC CGL 2024 final Result Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: 64
Solution : Given: $(x–4)(x^2+4x+16)=x^3–p$ Formula: We know that $a^3–b^3=(a–b)(a^2+ab+b^2)$ Solution: $x^3–p=(x–4)(x^2+4x+16)$ ⇒ $x^3−p=(x^3−4^3)$ ⇒ $p = 4^3$ So, $p = 64$. Hence, the correct answer is 64.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : A complete factorisation of $(x^4+64)$ is:
Option 1: $(x^2+8)^2$
Option 2: $(x^2+8)$$(x^2-8)$
Option 3: $(x^2-4x+8)(x^2-4x-8)$
Option 4: $(x^2+4x+8)$$(x^2-4x+8)$
Question : If $x$,$y$ and $z$ are real numbers such that $(x-3)^2+(y-4)^2+(z-5)^2=0$, then $(x+y+z)$ is equal to:
Option 1: –12
Option 2: 0
Option 3: 8
Option 4: 12
Question : If $x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{7+4 \sqrt{3}}}}$ where $x > 0$, then the value of $x$ is equal to:
Option 1: 3
Option 2: 4
Option 3: 1
Option 4: 2
Question : If $x+y=1+xy$, then $x^3+y^3-x^3y^3$ is equal to:
Option 1: 0
Option 2: 1
Option 3: –1
Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-2x+4=0$, then what is the equation whose roots are $\frac{\alpha ^{3}}{\beta ^{2}}$ and $\frac{\beta ^{3}}{\alpha ^{2}}?$
Option 1: $x^{2}-4x+8=0$
Option 2: $x^{2}-32x+4=0$
Option 3: $x^{2}-2x+4=0$
Option 4: $x^{2}-16x+4=0$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile