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:
Option 1: $\frac{1}{9}$
Option 2: $\frac{1}{3}$
Option 3: $\frac{1}{4}$
Option 4: $\frac{1}{2}$
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: $\frac{1}{4}$
Solution : $\frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}$ We know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ Let $a=p-q$, $b=q-r$ and $c=r-p$ We observe that $a+b+c=p-q+q-r+-p=0$ So, $a^3+b^3+c^3-3abc=0$ or, $a^3+b^3+c^3=3abc$ or, $(p-q)^3+(q-r)^3+(r-p)^3=3(p-q)(q-r)(r-p)$ So, $\frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}= \frac{3(p-q)(q-r)(r-p)}{12(p-q)(q-r)(r-p)}$ = $\frac{1}{4}$ Hence, the correct answer is $\frac{1}{4}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$, where $x \neq y \neq z$, is equal to:
Question : If $p=9, q=\sqrt{17}$, then the value of $(p^2-q^2)^{-\frac{1}{3}}$ is equal to:
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 $p^2+q^2=7pq$, then the value of $\frac{p}{q}+\frac{q}{p}$ is equal to:
Question : If $a+b=2c$, then the value of $\frac{a}{a–c}+\frac{c}{b–c}$ is equal to (where $a\neq b\neq c$):
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile