Question : The numerical value of $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$ is: $(a\neq b\neq c)$
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{3}$
Option 4: $3$
Correct Answer: $3$
Solution : Given: $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$ Taking LCM of the given expression, we have, = $\frac{(a–b)(a–b)^{2}+(b–c)(b–c)^{2}+(c–a)(c–a)^{2}}{(a–b)(b–c)(c–a)}$ = $\frac{(a–b)^{3}+(b–c)^{3}+(c–a)^{3}}{(a–b)(b–c)(c–a)}$ We know that $x^{3}+y^{3}+z^{3}=3xyz$, if $x+y+z=0$ According to this, we have, ⇒ $(a-b)=x$, $(b-c)=y$, $(c-a)=z$ So, $x+y+z=(a-b+b-c+c-a)=0$ Now, $\frac{(a–b)^{3}+(b–c)^{3}+(c–a)^{3}}{(a–b)(b–c)(c–a)}$ = $\frac{3(a–b)(b–c)(c–a)}{(a–b)(b–c)(c–a)}$ = 3 Hence, the correct answer is $3$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : If the sum of $\frac{a}{b}$ and its reciprocal is 1 and $a\neq 0,b\neq 0$, then the value of $a^{3}+b^{3}$ is:
Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}(a\neq b\neq c)$, then the value of $abc$ is:
Question : If $x=\frac{8ab}{a+b}(a\neq b),$ then the value of $\frac{x+4a}{x–4a}+\frac{x+4b}{x–4b}$ is:
Question : If $a+b+c=0$, then the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}$ is:
Question : If $(a^2 = b + c)$, $(b^2 = a + c)$, $(c^2 = b + a)$. Then, what will be the value of $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile