Question : If $\left (2a-1 \right )^{2}+\left (4b-3 \right)^{2}+\left (4c+5 \right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}$ is:
Option 1: $1\tfrac{3}{8}$
Option 2: $2\tfrac{3}{8}$
Option 3: $3\tfrac{3}{8}$
Option 4: $0$
New: SSC CHSL Tier 2 answer key released | 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: $0$
Solution : Given: $(2a-1)^{2}+(4b-3)^{2}+(4c+5)^{2}=0$ ⇒ $ 2a-1 = 0$, $4b-3$ and $4c+5=0$ [If the sum of squares is equal to zero then the individual term will also be zero] ⇒ $a=\frac{1}{2}$, $b=\frac{3}{4}$, and $c=\frac{-5}{4}$ $\therefore \small \frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}$ = $\frac{(\frac{1}{2})^3 +(\frac{3}{4})^3 + (\frac{-5}{4})^3 - 3\times \frac{1}{2}\times \frac{3}{4}\times\frac{-5}{4}}{(\frac{1}{2})^2 +(\frac{3}{4})^2 + (\frac{-5}{4})^2}$ = $\frac{\frac{1}{8}+\frac{27}{64}-\frac{125}{64}+\frac{45}{32}}{\frac{1}{4}+\frac{9}{16}+\frac{25}{16}}$ = $\frac{\frac{8+27-125+90}{64}}{\frac{4+9+25}{16}}$ = $0$ Hence, the correct answer is $0$.
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 $\small x=a\left (b-c \right),\; y=b\left (c-a \right) ,\; z=c\left (a-b \right)$, then the value of $\left (\frac{x}{a} \right)^{3}+\left (\frac{y}{b} \right)^{3}+\left (\frac{z}{c} \right)^{3}$ is:
Question : If $\left (a+b \right):\left (b+c \right):\left (c+a \right)= 6:7:8$ and $\left (a+b+c \right) = 14,$ then value of $c$ is:
Question : If $\small c+\frac{1}{c}=3$, then the value of $\left (c-3 \right )^{7}+\frac{1}{c^{7}}$ is:
Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Question : If $(a+b+c) \neq 0$, then $(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile