Question : If $p = 8.15,$ $q = 9.06$, and $r =-17.21$, then the value of $p^3+q^3+r^3-3 p q r$ is:
Option 1: −3.81
Option 2: −5.62
Option 3: 4.75
Option 4: 0
Latest: SSC CGL Tier 1 Result 2024 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: 0
Solution : $p = 8.15,$ $q = 9.06$ and $r =-17.21$ We know that, If $a+b+c=0$, then $a^3+b^3+c^3-3abc=0$ Here, $p+q+r=8.15+9.06+(-17.21)=0$ So, $p^3+q^3+r^3-3pqr=0$ Hence, the correct answer is 0.
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : If $x+y+z = 9$, then the value of $(x−4)^3+(y−2)^3+(z−3)^3−3(x−4)(y−2)(z−3)$ is:
Question : The value of $\frac{p^{2}- (q - r)^{2}}{(p + r)^{2} - (q)^{2}}$ + $\frac{q^{2} - (p - r)^{2}}{(p + q)^{2} - (r)^{2}}$ + $\frac{r^{2}- (p - q)^{2}}{(q + r)^{2} - (p)^{2}}$ is:
Question : If $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}$, find the value of $(pa+qb+rc)$.
Question : The value of $ \frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}$, where $p \neq q \neq r$, is equal to:
Question : If $\frac{1}{p}+\frac{1}{q}=\frac{1}{p+q}$, the value of $\left (p^{3}-q^{3}\right )$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile