Question : If $(p-q)=6,(r-q)=5$ and $(r-p)=3$, then find the value of $\frac{p^3+q^3+r^3-3 p q r}{p+q+r}$.
Option 1: 35
Option 2: 45
Option 3: 30
Option 4: 40
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 35
Solution : Given, $(p - q) = 6,$ $(r - q) = 5$ and $(r - p) = 3$ We know, $p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2+ r^2 - pq - pr - qr)$ $⇒\frac{p^3+q^3+r^3-3 p q r}{p+q+r} =\frac{(p + q + r)(p^2 + q^2+ r^2 - pq - pr - qr)}{p+q+r}$ $⇒\frac{p^3+q^3+r^3-3 p q r}{p+q+r} =p^2+q^2+r^2-pq-pr-qr$ $=\frac{1}{2}(2p^2+2q^2+2r^2-2pq-2pr-2qr)$ $=\frac{1}{2}[(p-q)^2+(q-r)^2+(r-p)^2]$ $=\frac{1}{2}[6^2+5^2+3^2]$ $=\frac{1}{2}[36+25+9]$ $=35$ Hence, the correct answer is 35.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $p+q+r=0$, then what is the simplified value of the expression $(\frac{p^2}{p^2-q r}+\frac{q^2}{q^2-p r}+\frac{r^2}{r^2-p q})$?
Question : If $6 \sec \theta=10$, then find the value of $\frac{5 \operatorname{cosec} \theta-3 \cot \theta}{4 \cos \theta+3 \sin \theta}$.
Question : If in a right-angled $\triangle P Q R$, $\tan Q=\frac{5}{12}$, then what is the value of $\cos Q$?
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile